# Hc Verma i Solutions for Class 11 Science Physics Chapter 8 - Work And Energy

Hc Verma i Solutions for Class 11 Science Physics Chapter 8 Work And Energy are provided here with simple step-by-step explanations. These solutions for Work And Energy are extremely popular among class 11 Science students for Physics Work And Energy Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma i Book of class 11 Science Physics Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Hc Verma i Solutions. All Hc Verma i Solutions for class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 130:

#### Question 1:

When you lift a box from the floor and put it on an almirah the potential energy of the box increases, but there is no change in its kinetic energy. Is it a violation of conservation of energy?

#### Answer:

No. Work done in lifting the box increases the potential energy of the box. During lifting at every point, the force applied by us on the box in the upward direction is equal to the gravitational force acting on the box in the downward direction. Therefore, there is no change in the velocity of the box. As a result, the kinetic energy of the box will not change.

#### Page No 130:

#### Question 2:

A particle is released from the top of an incline of height *h*. Does the kinetic energy of the particle at the bottom of the incline depend on the angle of incline? Do you need any more information to answer this question in Yes or No?

#### Answer:

No, the kinetic energy of the particle at the bottom of the inclined plane does not depend on the angle of inclination. When the particle reaches the ground, all its potential energy, while at the top of the inclined plane, is converted into kinetic energy. As we know that kinetic energy depends only on the height of the particle, it will be the same for different angles of inclination.

No, we do not need any other information to answer this question.

#### Page No 130:

#### Question 3:

Can the work by kinetic friction on an object be positive? Zero?

#### Answer:

Yes. Let us consider a block A which is resting on another block B. Block B is resting on a smooth horizontal surface. Let the coefficient of friction between the blocks be $\mathrm{\mu}$.

When a force *F *is applied on block B in the forward direction as shown in the above figure, block A moves with block B in the direction of the applied force. The frictional force on block A and the displacement will be in the forward direction. Therefore, work done by the frictional force is positive.

If we consider the reference frame of block B, then displacement of block A will be zero. Therefore, work done by the frictional force is zero.

#### Page No 130:

#### Question 4:

Can static friction do nonzero work on an object? If yes, give an example. If no, give reason.

#### Answer:

Yes. Let us consider a block A which is resting on another block B. Block B is resting on a smooth horizontal surface. Let the coefficient of kinetic friction between the blocks be ${\mathrm{\mu}}_{\mathrm{k}}$.

When a force *F *is applied on block B in the forward direction as shown in the above figure, block A moves with block B in the direction of the applied force. The friction force on block A and the displacement will be in the forward direction. Therefore, work done by the friction force is positive. In this case, block A will remain in contact with block B. This shows that static friction is doing a nonzero work on an object.

#### Page No 130:

#### Question 5:

Can normal force do nonzero work on an object? If yes, give an example. If no, give reason.

#### Answer:

Yes. Let us consider an elevator accelerating upward with a body placed in it. In this case, the normal reaction offered by the floor of the elevator on the body is greater than the weight of the body acting in the downward direction. If a person is observing this from the ground, then, for him, the normal reaction is doing a positive work, as the elevator is moving upward.

#### Page No 130:

#### Question 6:

Can kinetic energy of a system be increased without applying any external force on the system?

#### Answer:

Yes. Let us consider an isolated system of two particles falling towards each other under their mutual gravitational force of attraction. Here, the net force on the system is zero, but the velocities of the particles keep on increasing. Also, the kinetic energy of the system is increased without applying any external force on it.

#### Page No 130:

#### Question 7:

Is work-energy theorem valid in non-inertial frames?

#### Answer:

In an non-inertial frame, pseudo force also comes into account. As we know that pseudo force does not exist, work-energy theorem is not valid in non-inertial frames.

#### Page No 130:

#### Question 8:

A heavy box is kept on a smooth inclined plane and is pushed up by a force *F* acting parallel to the plane. Does the work done by the force *F* as the box goes from *A* to *B* depend on how fast the box was moving at *A* and *B*? Does the work by the force of gravity depend on this?

#### Answer:

(i) No. As the surface is smooth and the friction is zero, work done by the force will only depend on the force and the displacement.

(ii) No, because gravitational force is a conservative force and work done by a conservative force will depend only on the force and the displacement.

#### Page No 130:

#### Question 9:

One person says that the potential energy of a particular book kept in an almirah is 20 J and the other says it is 30 J. One of them necessarily wrong?

#### Answer:

No, both are correct. We measure potential energy from a reference level chosen by the observer. Therefore, in this case, both observers are measuring the potential energy from different reference levels.

#### Page No 130:

#### Question 10:

A book is lifted from the floor and is kept in an almirah. One person says that the potential energy of the book is increased by 20 J and the other says it is increased by 30 J. Is one of them necessarily wrong?

#### Answer:

Yes, one of them is necessarily wrong. We measure potential energy from a reference level chosen by the observer. However, the change in potential energy of a body does not depend on the level of reference.

#### Page No 130:

#### Question 11:

In one of the exercises to strengthen the wrist and fingers, a person squeezes and releases a soft rubber ball. Is the work done on the ball positive, negative or zero during compression? During expansion?

#### Answer:

(i) During compression, the work done on the ball is positive as the direction of the force applied by the fingers is along the compression of the ball.

(ii) During expansion, the work done is negative as expansion takes place against the force applied by the fingers on the ball.

#### Page No 130:

#### Question 12:

In tug of war, the team that exerts a larger tangential force on the ground wins. Consider the period in which a team is dragging the opposite team by applying a larger tangential force on the ground. List which of the following works are positive, which are negative and which are zero?

(a) work by the winning team on the losing team

(b) work by the losing team on the winning team

(c) work by the ground on the winning team

(d) work by the ground on the losing team

(e) total external work on the two teams.

#### Answer:

(a) Work by the winning team on the losing team is positive, as the displacement of the losing team is along the force applied by the winning team.

(b) Work by the losing team on the winning team is negative, as the displacement of the winning team is opposite to the force applied by losing team.

(c) Work by the ground on the winning team is positive.

(d) Work by the ground on the losing team is negative.

(e) Total external work on the two teams is positive.

#### Page No 131:

#### Question 13:

When an apple falls from a tree what happens to its gravitational potential energy just as it reaches the ground? After it strikes the ground?

#### Answer:

When an apple falls from a tree, its gravitational potential energy decreases as it reaches the ground. After it strikes the ground, its potential energy will remain unchanged.

#### Page No 131:

#### Question 14:

When you push your bicycle up on an inclined plane, the potential energy of the bicycle and yourself increases. Where does this energy come from?

#### Answer:

When a person pushes his bicycle up on an inclined plane, the potential energies of the bicycle and the person increase because moving up on the inclined plane the kinetic energy decreases. and as mechanical energy is sum of kinetic energy and potential energy, and remains constant for a conservative system. Therefore, potential energy must increase in this case.

#### Page No 131:

#### Question 15:

The magnetic force on a charged particle is always perpendicular to its velocity. Can the magnetic force change the velocity of the particles? Speed of the particle?

#### Answer:

The magnetic force on a charged particle is always perpendicular to its velocity. Therefore, the work done by the magnetic force on the charged particle is zero. Here, the kinetic energy and speed of the particle remain unaffected, while the velocity changes due to the change in direction of its motion.

#### Page No 131:

#### Question 16:

A ball is given a speed *v* on a rough horizontal surface. The ball travels through a distance *l* on the surface and stops. (a) what are the initial and final kinetic energies of the ball? (b) What is the work done by the kinetic friction?

#### Answer:

(a)

Initial kinetic energy of the ball, ${K}_{i}=\frac{1}{2}m{v}^{2}$

Here, *m* is the mass of the ball.

The final kinetic of the ball is zero.

(b)

Work done by the kinetic friction is equal to the change in kinetic energy of the ball.

∴ Work done by the kinetic friction = ${K}_{f}-{K}_{i}=0-\frac{1}{2}m{v}^{2}$

= $-\frac{1}{2}m{v}^{2}$

#### Page No 131:

#### Question 17:

Consider the situation of the previous question from a frame moving with a speed *v*_{0} parallel to the initial velocity of the block. (a) What are the initial and final kinetic energies? (b) What is the work done by the kinetic friction?

#### Answer:

The relative velocity of the ball w.r.t. the moving frame is given by ${v}_{r}=v-{v}_{0}$.

(a) Initial kinetic energy of the ball = $\frac{1}{2}m{{v}_{r}}^{2}=\frac{1}{2}m(v-{v}_{0}{)}^{2}$

Also, final kinetic energy of the ball = $\frac{1}{2}m(0-{v}_{0}{)}^{2}=\frac{1}{2}m{{v}_{0}}^{2}$

(b) Work done by the kinetic friction = final kinetic energy $-$ initial kinetic energy

= $\frac{1}{2}m({v}_{0}{)}^{2}-\frac{1}{2}m(v-{v}_{0}{)}^{2}$

= $-\frac{1}{2}m{v}^{2}+mv{v}_{0}$

#### Page No 131:

#### Question 1:

A heavy stone is thrown in from a cliff of height *h* in a given direction. The speed with which it hits the ground

(a) must depend on the speed of projection

(b) must be larger than the speed of projection

(c) must be independent of the speed of projection

(d) may be smaller than the speed of projection.

#### Answer:

(a) must depend on the speed of projection

(b) must be larger than the speed of projection

Consider that the stone is projected with initial speed *v*.

As the stone is falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.

From the conservation of energy, we have:

Initial energy of the stone = final energy of the stone

$\mathrm{i}.\mathrm{e}.,(K.E.{)}_{i}+(P.E.{)}_{i}=(K.E.{)}_{f}+(P.E.{)}_{f}$

$\Rightarrow \frac{1}{2}m{v}^{2}+mgh=\frac{1}{2}m({v}_{max}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{max}=\sqrt{{v}^{2}+2gh}\phantom{\rule{0ex}{0ex}}$

From the above expression, we can say that the maximum speed with which the stone hits the ground depends on the speed of projection and greater than it.

#### Page No 131:

#### Question 2:

The total work done on a particle is equal to the change in its kinetic energy

(a) always

(b) only if the forces acting on it are conservative

(c) only if gravitational force alone acts on it

(d) only if elastic force alone acts on it.

#### Answer:

(a) always

According to the work-energy theorem, the total work done on a particle is equal to the change in kinetic energy of the particle.

#### Page No 131:

#### Question 3:

Two equal masses are attached to the two ends of a spring of spring constant *k*. The masses are pulled out symmetrically to stretch the spring by a length *x* over its natural length. The work done by the spring on each mass is

(a) $\frac{1}{2}k{x}^{2}$

(b) $-\frac{1}{2}k{x}^{2}$

(c) $\frac{1}{4}k{x}^{2}$

(d) $-\frac{1}{4}k{x}^{2}$

#### Answer:

(d) $-\frac{1}{4}k{x}^{2}$

The work done by the spring on both the masses is equal to the negative of the increase in the elastic potential energy of the spring.

The elastic potential energy of the spring is given by ${E}_{p}=\frac{1}{2}k{x}^{2}$.

Work done by the spring on both the masses = $-\frac{1}{2}k{x}^{2}$

∴ Work done by the spring on each mass = $\frac{1}{2}\left(-\frac{1}{2}k{x}^{2}\right)=-\frac{1}{4}k{x}^{2}$

#### Page No 131:

#### Question 4:

The negative of the work done by the conservative internal forces on a system equal the changes in

(a) total energy

(b) kinetic energy

(c) potential energy

(d) none of these.

#### Answer:

(c) potential energy

The negative of the work done by the conservative internal forces on a system is equal to the changes in potential energy.

i.e. $W=-\u2206P.E.$

#### Page No 131:

#### Question 5:

The work done by the external forces on a system equals the change in

(a) total energy

(b) kinetic energy

(c) potential energy

(d) none of these

#### Answer:

(a) total energy

When work is done by an external forces on a system, the total energy of the system will change.

#### Page No 131:

#### Question 6:

The work done by all the forces (external and internal) on a system equals the change in

(a) total energy

(b) kinetic energy

(c) potential energy

(d) none of these

#### Answer:

(a) total energy

The work done by all the forces (external and internal) on a system is equal to the change in the total energy.

#### Page No 131:

#### Question 7:

____________ of a two particle system depends only on the separation between the two particles. The most appropriate choice for the blank space in the above sentence is

(a) Kinetic energy

(b) Total mechanical energy

(c) Potential energy

(d) Total energy.

#### Answer:

(c) Potential energy

The potential energy of a two particle system depends only on the separation between the particles.

#### Page No 131:

#### Question 8:

A small block of mass *m* is kept on a rough inclined surface of inclination θ fixed in an elevator. the elevator goes up with a uniform velocity *v* and the block does not slide on the wedge. The work done by the force of friction on the block in time *t* will be

(a) zero

(b) *mgvt* cos^{2}θ

(c) *mgvt* sin^{2}θ

(d) *mgvt* sin 2θ

#### Answer:

(c) *mgvt* sin^{2}θ

Distance (*d*) travelled by the elevator in time *t* *= vt*

The block is not sliding on the wedge.

Then friction force (*f*) = *mg* sin$\theta $

Work done by the friction force on the block in time *t* is given by

$W=Fd\mathrm{cos}(90-\theta )\phantom{\rule{0ex}{0ex}}\Rightarrow W=mg\mathrm{sin}\theta \times d\times \mathrm{cos}(90-\theta )\phantom{\rule{0ex}{0ex}}\Rightarrow W=mgd{\mathrm{sin}}^{2}\theta \phantom{\rule{0ex}{0ex}}\therefore W=mgvt{\mathrm{sin}}^{2}\theta $

#### Page No 131:

#### Question 9:

A block of mass *m* slides down a smooth vertical circular track. During the motion, the block is in

(a) vertical equilibrium

(b) horizontal equilibrium

(c) radial equilibrium

(d) none of these.

#### Answer:

(d) none of these.

The net force on the block is not zero, therefore the block will not be in any given equilibrium.

#### Page No 131:

#### Question 10:

A particle is rotated in a vertical circle by connecting it to a string of length *l* and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is

(a) $\sqrt{gl}$

(b) $\sqrt{2gl}$

(c) $\sqrt{3gl}$

(d) $\sqrt{5gl}$

#### Answer:

(c) $\sqrt{3gl}$

Suppose that one end of an extensible string is attached to a mass *m*, while the other end is fixed. The mass moves with a velocity *v* in a vertical circle of radius *R*. At some instant, the string makes an angle *θ* with the vertical as shown in the figure.

For a complete circle, the minimum velocity at L must be ${v}_{\mathrm{L}}=\sqrt{5gl}$.

Applying the law of conservation of energy, we have:

Total energy at M = total energy at L

$\mathrm{i}.\mathrm{e}.,\frac{1}{2}m{{v}_{\mathrm{M}}}^{2}+mgl=\frac{1}{2}m{{v}_{\mathrm{L}}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}m{{v}_{\mathrm{M}}}^{2}=\frac{1}{2}m{{v}_{\mathrm{L}}}^{2}-mgl\phantom{\rule{0ex}{0ex}}\mathrm{Using}{v}_{L}\ge \sqrt{5gl},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{{v}_{\mathrm{M}}}^{2}\ge \frac{1}{2}m\left(5gl\right)-mgl\phantom{\rule{0ex}{0ex}}\therefore {v}_{\mathrm{M}}=\sqrt{3gl}$

#### Page No 132:

#### Question 3:

A box is pushed through 4.0 m across a floor offering 100 N resistance. How much work is done by the resisting force?

#### Answer:

Resisting force acting on the box, $F=100\mathrm{N}$

Displacement of the box, *S = 4 *m

Also,$\theta =180\xb0\phantom{\rule{0ex}{0ex}}$

∴ Work done by the resisting force, $\phantom{\rule{0ex}{0ex}}\mathrm{W}=\underset{\mathrm{F}}{\to}\xb7\underset{\mathrm{S}}{\to}=100\times 4\times \mathrm{cos}180\xb0=-400\mathrm{J}$

#### Page No 132:

#### Question 4:

A block of mass 5.0 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.

#### Answer:

$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},M=5\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Angle}\mathrm{of}\mathrm{inclination},\theta =30\xb0$

Gravitational force acting on the block, $\mathrm{F}=mg$

Work done by the force of gravity depends only on the height of the object, not on the path length covered by the object.

$\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{object},h=10\times \mathrm{sin}30\xb0\phantom{\rule{0ex}{0ex}}=10\times \frac{1}{2}=5\mathrm{m}$

$\therefore \mathrm{Work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{force}\mathrm{of}\mathrm{gravity},w=mgh\phantom{\rule{0ex}{0ex}}=5\times 9.8\times 5=245\mathrm{J}$

#### Page No 132:

#### Question 5:

A constant force of 2⋅5 N accelerates a stationary particle of mass 15 g through a displacement of 2⋅5 m. Find the work done and the average power delivered.

#### Answer:

Given:

$F=2.50\mathrm{N},S=2.5\mathrm{m}\mathrm{and}m=15g=0.015\mathrm{kg}$

Work done by the force,

$W=F\mathit{\xb7}S\mathrm{cos}0\xb0\left(\mathrm{acting}\mathrm{along}\mathrm{the}\mathrm{same}\mathrm{line}\right)\phantom{\rule{0ex}{0ex}}=2.5\times 2.5=6.25\mathrm{J}$

Acceleration of the particle is,

$a=\frac{F}{m}=\frac{2.5}{0.015}\phantom{\rule{0ex}{0ex}}=\frac{500}{3}\mathrm{m}/{\mathrm{s}}^{2}$

Applying the work-energy principle for finding the final velocity of the particle,

$\frac{1}{2}m{v}^{2}-0=6.25\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =\sqrt{\frac{6.25\times 2}{0.15}}=28.86\mathrm{m}/s$

So, time taken by the particle to cover 2.5 m distance,

$t=\frac{\nu -u}{\alpha}=\frac{\left(28.86\right)\times 3}{500}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Average}\mathrm{power}=\frac{W}{t}\phantom{\rule{0ex}{0ex}}=\frac{6.25\times 500}{\left(28.86\right)\times 3}=36.1\mathrm{W}$

#### Page No 132:

#### Question 6:

A particle moves from a point ${\underset{r}{\to}}_{1}=\left(2\mathrm{m}\right)\underset{i}{\to}+\left(3\mathrm{m}\right)\underset{j}{\to}$ to another point ${\underset{r}{\to}}_{2}=\left(3\mathrm{m}\right)\underset{i}{\to}+\left(2\mathrm{m}\right)\underset{j}{\to}$ acts on it. Find the work done by the force on the particle during the displacement.

#### Answer:

Initial position vector,

$\overrightarrow{{r}_{1}}=2\overrightarrow{i}+3\overrightarrow{j}$

Final position vector,

${\overrightarrow{r}}_{2}=3\overrightarrow{i}+2\overrightarrow{j}$

So, displacement vector,

$\overrightarrow{r}={\overrightarrow{r}}_{2}-{\overrightarrow{r}}_{1}\phantom{\rule{0ex}{0ex}}=\left(3\overrightarrow{i}+2\overrightarrow{j}\right)-\left(2\overrightarrow{i}+\overrightarrow{j}\right)\phantom{\rule{0ex}{0ex}}=\overrightarrow{i}-\overrightarrow{j}\phantom{\rule{0ex}{0ex}}\mathrm{Force}\mathrm{acting}\mathrm{on}\mathrm{the}\mathrm{particle},\overrightarrow{\mathit{F}}=5\overrightarrow{i}+5\overrightarrow{j}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{work}\mathrm{done}=\overrightarrow{\mathit{F}}\mathit{\xb7}\overrightarrow{\mathit{S}}=5\times 1+5\left(-1\right)=0$

#### Page No 132:

#### Question 7:

A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he covers a distance of 40 m with an acceleration of 0⋅5 m/s^{2}, find the work done by the man on the block during the motion.

#### Answer:

Given:

$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=2\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{coverd}\mathrm{by}\mathrm{the}\mathrm{man},s=40\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}\mathrm{of}\mathrm{the}\mathrm{man},\mathrm{a}=0.5\mathrm{m}/{s}^{2}$

So, force applied by the man on the box,

$F=ma\phantom{\rule{0ex}{0ex}}=2\times \left(0.5\right)\phantom{\rule{0ex}{0ex}}=1\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{Work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{man}\mathrm{on}\mathrm{the}\mathrm{block},W=F\mathit{\xb7}S\phantom{\rule{0ex}{0ex}}=1\times 40\phantom{\rule{0ex}{0ex}}=40\mathrm{J}$

#### Page No 132:

#### Question 8:

The kinetic energy force on the particle continuously increases with time.

(a) The resultant force on the particle must be parallel to the velocity at all instants.

(b) The resultant force on the particle must be at an angle less than 90° all the time.

(c) Its height above the ground level must continuously decrease.

(d) The magnitude of its linear momentum is increasing continuously.

#### Answer:

(b) The resultant force on the particle must be at an angle less than 90° with the velocity all the time.

(d) The magnitude of its linear momentum is increasing continuously.

Kinetic energy of a particle is directly proportional to the square of its velocity. The resultant force on the particle must be at an angle less than 90° with the velocity all the time so that the velocity or kinetic energy of the particle keeps on increasing.

The kinetic energy is also directly proportional to the square of its momentum, therefore it continuously increases with the increase in momentum of the particle.

#### Page No 132:

#### Question 9:

One end of a light spring of spring constant *k* is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is $\frac{1}{2}k{x}^{2}$. The possible cases are

(a) at spring was initially compressed by a distance *x* and was finally in its natural length

(b) it was initially stretched by a distance *x* and and finally was in its natural length

(c) it was initially in its natural length and finally in a compressed position

(d) it was initially in its natural length and finally in a stretched position.

#### Answer:

(a) at spring was initially compressed by a distance *x* and was finally in its natural length

(b) it was initially stretched by a distance *x* and and finally was in its natural length

For an elastic spring, the work done is equal to the negative of the change in its potential energy.

When the spring was initially compressed or stretched by a distance *x*, its potential energy is given by

*${\left(P.E.\right)}_{i}=\frac{1}{2}k{x}^{2}$*.

When it finally comes to its natural length, its potential energy is given by

*${\left(P.E.\right)}_{f}=0$*.

∴ Work done = *$-\left[{\left(P.E.\right)}_{f}-{\left(P.E.\right)}_{i}\right]=-\left[0-\frac{1}{2}k{x}^{2}\right]=\frac{1}{2}k{x}^{2}$*

#### Page No 132:

#### Question 10:

A block of mass *M* is hanging over a smooth and light pulley through a light string. T he other end of the string is pulled by a constant force *F*. The kinetic energy of the block increases by 20 J in 1 s.

(a) The tension in the string is *Mg*.

(b) The tension in the string is *F*.

(c) The work one by the tension on the block is 20 J in the above 1 s.

(d) The work done by the force of gravity is −20 J in the above 1 s.

#### Answer:

(b) The tension in the string is *F*.

Tension in the string is equal to* F*, as tension on both sides of a frictionless and massless pulley is the same.

i.e., *T – Mg = Ma
$\Rightarrow $T = Mg + Ma*

So, the tension in the string cannot be equal to

*Mg*.

The change in kinetic energy of the block is equal to the work done by gravity.

Hence, the work done by gravity is 20 J in 1 s, while the the work done by the tension force is zero.

#### Page No 132:

#### Question 1:

The mass of a cyclist together with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6⋅0 km/h to 12 km/h.

#### Answer:

Total mass of the system (cyclist and bike), $M={m}_{c}+{m}_{b}=90\mathrm{kg}$

Initial velocity of the system, $u=6.0\mathrm{km}/\mathrm{h}=1.666\mathrm{m}/\mathrm{sec}$

Final velocity of the system, $\nu =12\mathrm{km}/\mathrm{h}=3.333\mathrm{m}/\mathrm{sec}$

From work-energy theorem, we have:

$\mathrm{Increase}\mathrm{in}\mathrm{K}.\mathrm{E}.=\frac{1}{2}M{\nu}^{2}-\frac{1}{2}m{u}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}90\times {\left(3.333\right)}^{2}-\frac{1}{2}\times 90\times {\left(1.66\right)}^{2}\phantom{\rule{0ex}{0ex}}=499.4-124.6\phantom{\rule{0ex}{0ex}}=374.8=375\mathrm{J}$

#### Page No 132:

#### Question 2:

A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.00 m/s^{2} for 5.00 s. Compute its final kinetic energy.

#### Answer:

$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},{M}_{b}=2\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{block},u=10\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\mathrm{Also},a=3\mathrm{m}/{s}^{2}\mathrm{and}t=5\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{motion},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\nu =u+at\phantom{\rule{0ex}{0ex}}=10+3\times 5=25\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Final}\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\nu}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 2\times 625=625\mathrm{J}$

#### Page No 133:

#### Question 8:

A force $F=\alpha +bx$ acts on a particle in the *x*-direction, where *a* and *b* are constants. Find the work done by this force during a displacement from *x* = 0 to *x* = d.

#### Answer:

Given that force is a function of displacement, i.e. $\mathrm{F}=a+bx$,

where *a* and *b* are constants.

So, work done by this force during the displacement *x* = 0 to *x* = *d*,

$\mathrm{W}=\underset{0}{\overset{d}{\int}}\mathrm{F}dx\phantom{\rule{0ex}{0ex}}W=\underset{0}{\overset{d}{\int}}\left(a+bx\right)dx\phantom{\rule{0ex}{0ex}}W={\left[ax+\frac{b{x}^{2}}{2}\right]}_{0}^{d}\phantom{\rule{0ex}{0ex}}W=ad+\frac{b{d}^{2}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow W=\left(a+\frac{bd}{2}\right)d$

#### Page No 133:

#### Question 9:

A block of mass 250 g slides down an incline of inclination 37° with uniform speed. Find the work done against friction as the block slides through 1m.

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}m=250g,\mathrm{\theta}=37\xb0,d=1\mathrm{m}$

Here, R is the normal reaction of the block.

As the block is moving with uniform speed,

$f=mg\mathrm{sin}37\xb0$

So, work done against the force of friction,

$W=fd\mathrm{cos}0\xb0\phantom{\rule{0ex}{0ex}}W=(mg\mathrm{sin}37\xb0)\times d\phantom{\rule{0ex}{0ex}}W=(0.25\times 9.8\times \mathrm{sin}37\xb0)\times 1.0\phantom{\rule{0ex}{0ex}}W=1.5\mathrm{J}$

#### Page No 133:

#### Question 10:

A block of mass *m* is kept over another block of mass *M* and the system rests on a horizontal surface (figure 8-E1). A constant horizontal force *F* acting on the lower block produces an acceleration $\frac{F}{2\left(m+M\right)}$ in the system, and the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement *d* of the system.

Figure

#### Answer:

(a)

$a=\frac{\mathrm{F}}{2\left(\mathrm{M}+m\right)}\left(\mathrm{given}\right)$

The free-body diagrams of both the blocks are shown below:

For the block of mass *m*,

$ma={\mu}_{1}{\mathrm{R}}_{1}\mathrm{and}{\mathrm{R}}_{1}=mg\phantom{\rule{0ex}{0ex}}\Rightarrow {\mu}_{1}=\frac{ma}{{\mathrm{R}}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{F}{2\left(\mathrm{M}+m\right)g}$

(b)

Frictional force acting on the smaller block,

${f}_{1}={\mu}_{1}{\mathrm{R}}_{1}=\frac{\mathrm{F}}{2\left(\mathrm{M}+m\right)g}\times mg\phantom{\rule{0ex}{0ex}}=\frac{m\mathrm{F}}{2(\mathrm{M}+m)}$

(c) Work done, *w* = *f*_{1}*s* [where *s* = *d*]

$=\frac{m\mathrm{F}}{2\left(\mathrm{M}+m\right)}\times d\phantom{\rule{0ex}{0ex}}=\frac{m\mathrm{F}d}{2\left(\mathrm{M}+m\right)}$

#### Page No 133:

#### Question 11:

A box weighing 2000 N is to be slowly slid through 20 m on a straight track with friction coefficient 0⋅2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ, which ensures him the minimum magnitude of the force.

#### Answer:

Given:

$\mathrm{Weight}=2000\mathrm{N},\mathrm{s}=20\mathrm{m},\mu =0.2$

The free-body diagram for the box is shown below:

(a) From the figure,

$R+P\mathrm{sin}\mathrm{\theta}-2000=0...\left(i\right)\phantom{\rule{0ex}{0ex}}P\mathrm{cos}\mathrm{\theta}-0.2R=0...\left(ii\right)$

From (i) and (ii),

$P\mathrm{cos}\mathrm{\theta}-0.2\left(2000-P\mathrm{sin}\mathrm{\theta}\right)=0\phantom{\rule{0ex}{0ex}}P\left(\mathrm{cos}\mathrm{\theta}+0.2\mathrm{sin}\mathrm{\theta}\right)=400\phantom{\rule{0ex}{0ex}}P=\frac{400}{\mathrm{cos}\mathrm{\theta}+0.2\mathrm{sin}\mathrm{\theta}}...\left(iii\right)$

So, work done by the person,

$W\mathit{=}PS\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=\frac{8000\mathrm{cos}\mathrm{\theta}}{\mathrm{cos}\mathrm{\theta}+0.2\mathrm{sin}\mathrm{\theta}}\phantom{\rule{0ex}{0ex}}=\frac{8000}{1+0.2\mathrm{tan}\mathrm{\theta}}\phantom{\rule{0ex}{0ex}}=\frac{40000}{5+\mathrm{tan}\mathrm{\theta}}...\left(iv\right)$

(b) For minimum magnitude of force from equation (*iii*),

$\frac{d}{dk}\left(\mathrm{cos}\mathrm{\theta}+0.2\mathrm{sin}\mathrm{\theta}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\mathrm{\theta}=0.2$

Putting the value in equation (*iv*),

$W=\frac{40000}{\left(5+\mathrm{tan}\mathrm{\theta}\right)}\phantom{\rule{0ex}{0ex}}=\frac{40000}{5+0.2}\approx 7692\mathrm{J}$

#### Page No 133:

#### Question 12:

A block of weight 100 N is slowly moved up a smooth incline of inclination 37° by a person. Calculate the work done by the person in moving the block through a distance of 2 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction.

#### Answer:

Given:

$Weight,mg=100\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{\theta}=37\xb0\mathrm{and}s=2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Force},\mathrm{F}=mg\mathrm{sin}37\xb0\phantom{\rule{0ex}{0ex}}=100\times \frac{3}{5}=60\mathrm{N}$

So, work done when the force is parallel to incline,

$W=FS\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=60\times 2\times \mathrm{cos}0\xb0=120\mathrm{J}$

$\mathrm{In}\Delta \mathrm{ABC},\mathrm{AB}=2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=h\phantom{\rule{0ex}{0ex}}=s\times \mathrm{sin}37\xb0=2.0\times \mathrm{sin}37\xb0\phantom{\rule{0ex}{0ex}}=1.2\mathrm{m}$

∴ Work done when the force is in horizontal direction,

$W\text{'}=mgh\phantom{\rule{0ex}{0ex}}=100\times 1.2=120\mathrm{J}$

#### Page No 133:

#### Question 13:

Find the average frictional force needed to stop a car weighing 500 kg at a distance of 25 m if the initial speed is 72 km/h.

#### Answer:

$\text{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{car},m=500\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{car},s=25\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{car},u=72\mathrm{km}/\mathrm{h}=20m/s\phantom{\rule{0ex}{0ex}}\mathrm{Final}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{car},\nu =0m/s$

Retardation of the car,

$a=\frac{\left({\nu}^{2}-{u}^{2}\right)}{2s}\phantom{\rule{0ex}{0ex}}\Rightarrow a=-\frac{400}{50}=-8m/{s}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Frictional}\mathrm{force},F=ma=500\times 8=4000\mathrm{N}$

#### Page No 133:

#### Question 14:

Find the average force needed to accelerate a car weighing 500 kg from rest to 72 km/h through a distance of 25 m.

#### Answer:

$\text{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{car},m=500\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{car},u=0\phantom{\rule{0ex}{0ex}}\mathrm{Final}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{car},\nu =72\mathrm{km}/\mathrm{h}=20\mathrm{m}/s\phantom{\rule{0ex}{0ex}}a=\frac{\left({\nu}^{2}-{u}^{2}\right)}{2s}\phantom{\rule{0ex}{0ex}}a=\frac{400}{50}=8\mathrm{m}/{s}^{2}$

Force needed to accelerate the car,

$\mathrm{F}=ma=500\times 8=400\mathrm{N}$

#### Page No 133:

#### Question 15:

A particle of mass *m* moves on a straight line with its velocity varying with the distance travelled, according to the equation $\nu =a\sqrt{x}$, where *a* is a constant. Find the total work done by all the forces during a displacement from $x=0\mathrm{to}x-d$.

#### Answer:

Given,

$\nu =a\sqrt{x}\left(\mathrm{uniformly}\mathrm{accelerated}\mathrm{motion}\right)\phantom{\rule{0ex}{0ex}}\text{D}\mathrm{isplacement},s=d-0=d\phantom{\rule{0ex}{0ex}}\text{P}\mathrm{utting}x=0,\text{weget}{\nu}_{1}=0\phantom{\rule{0ex}{0ex}}\text{P}\mathrm{utting}x=d,\text{weget}{\nu}_{2}=a\sqrt{d}\phantom{\rule{0ex}{0ex}}\alpha =\frac{{\nu}_{2}^{2}-{\nu}_{1}^{2}}{2s}=\frac{{a}^{2}d}{2d}=\frac{{a}^{2}}{2}$

$\mathrm{Force},F=m\alpha =\frac{m{a}^{2}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Work}\mathrm{done},\mathrm{W}=Fs\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=\frac{m{a}^{2}}{2}\times d=\frac{m{a}^{2}d}{2}$

#### Page No 133:

#### Question 16:

A block of mass 2 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take g = 10 m/s^{2}.

#### Answer:

(a)

Given:

$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=2\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{\theta}=37\xb0\phantom{\rule{0ex}{0ex}}\mathrm{Force}\mathrm{on}\mathrm{the}\mathrm{block},F=20\mathrm{N}$

From the above figure,

$F=\left(2g\mathrm{sin}\mathrm{\theta}\right)+ma\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{20-20\mathrm{sin}\mathrm{\theta}}{2}=4m/{s}^{2}\phantom{\rule{0ex}{0ex}}s=ut+\frac{1}{2}a{t}^{2}=2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{work}\mathrm{done}\phantom{\rule{0ex}{0ex}}W=FS=20\times 2=40\mathrm{J}$

(b) If $W=40\mathrm{J}$

$S\mathit{=}\frac{\mathit{W}}{\mathit{F}}=\frac{40}{20}=2\mathrm{m}\phantom{\rule{0ex}{0ex}}h=2\mathrm{sin}37\xb0=1.2\mathrm{m}$

So, work done

$\mathrm{W}=-mgh\phantom{\rule{0ex}{0ex}}=-20\times 1.2=-24\mathrm{J}$

(c)

$\nu =u+at\phantom{\rule{0ex}{0ex}}=4\times 1=4\mathrm{m}/\mathrm{sec}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 2\times 16=16\mathrm{J}$

#### Page No 133:

#### Answer:

Given:

$\mathrm{Mass},m=2\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Inclination},\mathrm{\theta}=37\xb0\phantom{\rule{0ex}{0ex}}\mathrm{Force}\mathrm{applied},\mathrm{F}=20\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}\mathrm{of}\mathrm{the}\mathrm{block},a=10\mathrm{m}/{s}^{2}$

(a) *t* = 1 sec

So, $s=ut+\frac{1}{2}a{t}^{2}=5\mathrm{m}$

Work done by the applied force,

$W\mathit{=}Fs\mathrm{cos}\theta \xb0=20\times 5=100\mathrm{J}$

(b) $\mathrm{AB}\left(\mathrm{h}\right)=5\mathrm{sin}37\xb0=3\mathrm{m}$

So, work done by weight,

$\mathrm{W}=mgh\phantom{\rule{0ex}{0ex}}2\times 10\times 3=60\mathrm{J}$

So, frictional force,

$f=mg\mathrm{sin}\mathrm{\theta}$

Work done by the friction forces,

$\mathrm{W}=fs\mathrm{cos}0\xb0=\left(mg\mathrm{sin}\mathrm{\theta}\right)s\phantom{\rule{0ex}{0ex}}=20\times 0.60\times 5=-60\mathrm{J}$

#### Page No 133:

#### Question 18:

A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 cm/s. If the friction coefficient between the table and the block is 0⋅1, how far does the block move before coming to rest?

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=250\mathrm{gm}=0.250\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{block},u=40\mathrm{cm}/s=0.4\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\mathrm{Final}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{block},\nu =0\phantom{\rule{0ex}{0ex}}\mathrm{Coefficient}\mathrm{of}\mathrm{friction},\mu =0.1$

Force in the forward direction is equal to the friction force.

$\mathrm{Here},\mu \mathrm{R}=ma\phantom{\rule{0ex}{0ex}}\left(\mathrm{where}a\mathrm{is}\mathrm{deceleration}\right)\phantom{\rule{0ex}{0ex}}a=\left(\frac{\mu \mathrm{R}}{m}\right)=\left(\frac{\mu mg}{m}\right)=\mu g\phantom{\rule{0ex}{0ex}}=0.1\times 9.8=0.98m/{s}^{2}\phantom{\rule{0ex}{0ex}}s=\frac{{\nu}^{2}-{u}^{2}}{2a}=0.082\mathrm{m}\phantom{\rule{0ex}{0ex}}=8.2\mathrm{cm}$

Again, work done against friction,

$\mathrm{W}=-\mu Rs\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=-1\times 2.5\times 0.082\times 1\phantom{\rule{0ex}{0ex}}=-0.02\mathrm{J}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{W}=-0.02\mathrm{J}$

#### Page No 133:

#### Question 19:

Water falling from a 50-m high fall is to be used for generating electric energy. If $1\xb78\times {10}^{5}\mathrm{kg}$ of water falls per hour and half the gravitational potential energy can be converted into electrical energy, how many 100 W lamps can be lit with the generated energy?

#### Answer:

Given:

$\mathrm{Height},h=50\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{water}\mathrm{falling}\mathrm{per}\mathrm{hour},m=1.8\times {10}^{5}\mathrm{kg}$

Power of a lamp,

$P=100\mathrm{watt}\phantom{\rule{0ex}{0ex}}\mathrm{Potential}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{water},\phantom{\rule{0ex}{0ex}}\mathrm{P}.\mathrm{E}.=mgh\phantom{\rule{0ex}{0ex}}=1.8\times {10}^{5}\times 9.8\times 50\phantom{\rule{0ex}{0ex}}=882\times {10}^{5}\mathrm{J}$

As only half the potential energy of water is converted into electrical energy,

$\mathrm{Electrical}\mathrm{energy}=\frac{1}{2}\mathrm{P}.\mathrm{E}.=441\times {10}^{5}\mathrm{J}/\mathrm{hr}$

So, power in watt $\left(\mathrm{J}/\mathrm{sec}\right)=\left(\frac{441\times {10}^{5}}{60\times 60}\right)$

Therefore, the number of 100 W lamps that can be lit using this energy,

$n=\frac{441\times {10}^{5}}{3600\times 100}=122.5\approx 122$

#### Page No 133:

#### Question 20:

A person is painting the walls of his house. He stands on a ladder with a bucket containing paint in one hand and a brush in the other. Suddenly the bucket slips from his hand and falls to the floor. If the bucket with the paint had a mass of 6 kg and was at a height of 2 m at the time it slipped, how much gravitational potential energy was lost together with the paint?

#### Answer:

Given:

$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{bucket}\mathrm{with}\mathrm{the}\mathrm{paint},m=6\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Height}\text{atwhichthebucketisplaced},h=2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Potential}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{b}\mathrm{u}\mathrm{c}\mathrm{k}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{p}\mathrm{a}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{at}\mathrm{the}\mathrm{given}\mathrm{height},\phantom{\rule{0ex}{0ex}}\mathrm{P}.\mathrm{E}.=mgh\phantom{\rule{0ex}{0ex}}=6\times \left(9.8\right)\times 2\phantom{\rule{0ex}{0ex}}=117.6\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{P}.\mathrm{E}.\text{onthe}\mathrm{floor}=0\phantom{\rule{0ex}{0ex}}\mathrm{Loss}\mathrm{in}\mathrm{potential}\mathrm{energy}=117.6-0\phantom{\rule{0ex}{0ex}}=117.6\mathrm{J}\approx 118\mathrm{J}$

#### Page No 133:

#### Question 21:

A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.

#### Answer:

Given:

Height of the cliff, *h* = 40 *m*

Initial speed of the projectile, *u* = 50 m/s

Let the projectile hit the ground with velocity '*v*'.

Applying the law of conservation of energy,

$mgh+\frac{1}{2}m{u}^{2}=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 10\times 40+\left(\frac{1}{2}\right)\times 2500=\frac{1}{2}{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}^{2}=3300\phantom{\rule{0ex}{0ex}}\Rightarrow v=57.4m/s=58\mathrm{m}/s$

The projectile hits the ground with a speed of 58 m/s.

#### Page No 133:

#### Question 22:

The 200 m free-style women's swimming gold medal at Seoul Olympics in 1988 was won by Heike Friendrich of East Germany when she set a new Olympic record of 1 minute and 57⋅56 seconds. Assume that she covered most of the distance with a uniform speed and had to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water.

#### Answer:

$\mathrm{Time}\mathrm{taken}\mathrm{to}\mathrm{cover}200\mathrm{m},t=1\mathrm{min}57.56\mathrm{seconds}=117.56\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Power}\mathrm{exerted}\mathrm{by}\mathrm{her},\phantom{\rule{0ex}{0ex}}P=460\mathrm{W}\phantom{\rule{0ex}{0ex}}P=\frac{\mathrm{W}}{t}\phantom{\rule{0ex}{0ex}}\mathrm{Work}done,\mathrm{W}=Pt=460\times 117.56\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{Again},\mathit{}W=Fs\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{W}{\mathrm{s}}\phantom{\rule{0ex}{0ex}}=\frac{460\times 117.56}{200}\phantom{\rule{0ex}{0ex}}=270.3\mathrm{N}\approx 270\mathrm{N}$

∴ Resistance force offered by the water during the swim is 270 N.

#### Page No 133:

#### Question 23:

The US athlete Florence Griffith-Joyner won the 100 m sprint gold medal at Seoul Olympics in 1988, setting a new Olympic record of 10⋅54 s. Assume that she achieved her maximum speed in a very short time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith-Joyner at her full speed. (b) Assuming that the track, wind etc. offered an average resistance of one-tenth of her weight, calculate the work done by the resistance during the run. (c) What power Griffith-Joyner had to exert to maintain uniform speed?

#### Answer:

Given:

Distance covered by her, *s *= 100 m

Time taken by her to cover 100 m, *t* = 10.54 s

Mass, *m* = 50 kg

The motion can be assumed to be uniform.

(a)

$\mathrm{Speed},\nu =\frac{\mathrm{s}}{t}=9.487\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{m\nu}}^{2}=2250\mathrm{J}$

(b)

$\mathrm{Weight}=mg=490\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{Average}\mathrm{resistance}\mathrm{force}\mathrm{offered},\phantom{\rule{0ex}{0ex}}R=\frac{mg}{10}=49\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{work}\mathrm{done}\mathrm{against}\mathrm{the}\mathrm{resitance}\mathrm{force}\phantom{\rule{0ex}{0ex}}\mathrm{W}=-Rs=-49\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{W}=-4900\mathrm{J}$

(c)

To maintain uniform speed, she had to exert 4900 J of energy to overcome friction.

Power exerted by her to overcome frcition,

$\mathrm{P}=\frac{\mathrm{W}}{t}\phantom{\rule{0ex}{0ex}}=\frac{4900}{10.54}=465\mathrm{W}$

#### Page No 133:

#### Question 24:

A water pump lifts water from 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower that the engine should have to do this.

#### Answer:

Given:

Height through which water is lifted,* h* = 10 m

$\mathrm{Flow}\mathrm{rate}\mathrm{of}\mathrm{water}=\left(\frac{m}{t}\right)\phantom{\rule{0ex}{0ex}}=30\mathrm{kg}/\mathrm{min}=0.5\mathrm{kg}/s$

Power delivered by the engine,

$P=\frac{mgh}{t}\phantom{\rule{0ex}{0ex}}=\left(0.5\right)\times 9.8\times 10\phantom{\rule{0ex}{0ex}}=49\mathrm{W}$

1 hp = 746 w

So, the minimum horse power (hp) that the engine should possess

$=\frac{p}{746}=\left(\frac{49}{746}\right)\phantom{\rule{0ex}{0ex}}=6.6\times {10}^{-2}\mathrm{hp}$

#### Page No 133:

#### Question 25:

An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3 m/s. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw it, what horsepower does he use?

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\text{stone},m=200g=0.2\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Height}\text{towhich}\mathrm{the}\mathrm{stone}\text{islifted},h=150\mathrm{cm}=1.5\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Velocity}\mathrm{of}\mathrm{the}\mathrm{projection},\nu =3\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Time},t=1\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Total}\mathrm{work}\mathrm{done},W=K.E.+P.E.\phantom{\rule{0ex}{0ex}}W=\frac{1}{2}m{\nu}^{2}+mgh\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)\times \left(0.2\right)\times 9+\left(0.2\right)\left(9.8\right)\times \left(1.5\right)\phantom{\rule{0ex}{0ex}}=3.84\mathrm{J}$

1 hp = 764 watt

Horsepower used by demonstrator

$=\frac{3.84}{746}=\left(5.14\right)\times {10}^{-3}$

Therefore, power used by the demonstrator to lift and throw the stone is 5.14$\times $10^{-3} hp.

#### Page No 133:

#### Question 26:

In a factory, 2000 kg of metal needs to be lifted by an engine through a distance of 12 m in 1 minute. Find the minimum horsepower of the engine to be used.

#### Answer:

Given:

Mass of the metal, $m=2000\mathrm{kg}$

Distance, *s* = 12 m

Time taken, *t* = 1 minute = 60 s

Force applied by the engine to lift the metal,

*F = mg*

$\mathrm{So},\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{engine},\phantom{\rule{0ex}{0ex}}W=F\times s\times \mathrm{cos}\mathrm{\theta}=mgs\times \mathrm{cos}0\xb0[\mathrm{\theta}=0\xb0\mathrm{for}\mathrm{minimum}\mathrm{force}]\phantom{\rule{0ex}{0ex}}=2000\times 10\times 12\phantom{\rule{0ex}{0ex}}=240000\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{power}\mathrm{exerted}\mathrm{by}\mathrm{the}\mathrm{engine},\phantom{\rule{0ex}{0ex}}P\mathit{=}\frac{W}{t}\phantom{\rule{0ex}{0ex}}=\frac{240000}{60}=4000\mathrm{watt}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Power}\mathrm{in}\mathrm{hp},\phantom{\rule{0ex}{0ex}}P=\frac{4000}{746}=5.3\mathrm{hp}$

#### Page No 133:

#### Question 27:

A scooter company gives the following specifications about its product:

Weight of the scooter − 95 kg

Maximum speed − 60 km/h

Maximum engine power − 3⋅5 hp

Pick up time to get the maximum speed − 5 s

Check the validity of these specifications.

#### Answer:

The specifications given by the company are:

$\mathrm{Mass},m=95\mathrm{kg}\phantom{\rule{0ex}{0ex}}Maximumpower,{P}_{m}=3.5\mathrm{hp}\phantom{\rule{0ex}{0ex}}Maximumspeed,{v}_{m}=60\mathrm{km}/\mathrm{h}\phantom{\rule{0ex}{0ex}}=\frac{50}{3}\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\mathrm{Pick}\mathrm{up}\mathrm{time}\mathrm{to}\mathrm{get}\mathrm{maximum}\mathrm{speed},{t}_{m}=5\mathrm{sec}$

So, the maximum acceleration that can be produced,

$a=\frac{50}{3\times 5}=\frac{10}{3}\mathrm{m}/{s}^{2}$

So, the driving force,

$F=ma=95\times \left(\frac{10}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{950}{3}\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{Max}\mathrm{speed},\nu =\frac{p}{F}\phantom{\rule{0ex}{0ex}}\Rightarrow v=3.5\times 746\times \frac{3}{950}\Rightarrow 8.2\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

As the scooter can reach a maximum of 8.2 m/s while producing a force of 950/3 N, the specifications given are not correct.

#### Page No 134:

#### Question 28:

A block of mass 30 kg is being brought down by a chain. If the block acquires a speed of 40 cm/s in dropping down 2 m, find the work done by the chain during the process.

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=30\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Speed}\mathrm{acquire}\text{d}\mathrm{by}\mathrm{the}\mathrm{block},\nu =40\mathrm{cm}/s\phantom{\rule{0ex}{0ex}}=0.4\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{block},s=2\mathrm{m}$

Let *a* be the acceleration of the block in the downward direction.

From the diagram, the force applied by the chain on the block,

$\mathrm{F}=\left(ma-mg\right)\phantom{\rule{0ex}{0ex}}=m\left(a-g\right)$

$a=\frac{{\nu}^{2}-{u}^{2}}{2\mathrm{s}}\phantom{\rule{0ex}{0ex}}=\frac{16}{-4}=0.04\mathrm{m}/{s}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{chain},\phantom{\rule{0ex}{0ex}}\mathrm{W}=Fs\mathrm{cos}\mathrm{\theta}$

$=m\left(a-g\right)\times \mathrm{s}\mathrm{cos}0\xb0\phantom{\rule{0ex}{0ex}}=30\left(0.04-9.8\right)\times 2\phantom{\rule{0ex}{0ex}}=-30\times \left(9.76\right)\times 2\phantom{\rule{0ex}{0ex}}=-585.6=-586\mathrm{J}\phantom{\rule{0ex}{0ex}}\Rightarrow W=-586\mathrm{J}$

#### Page No 134:

#### Question 29:

The heavier block in an Atwood machine has mass twice that of the lighter one. The tension in the string is 16⋅0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Tension}\mathrm{in}\mathrm{the}\mathrm{string},T=16\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{the}\mathrm{free}-\mathrm{body}\mathrm{diagrams},\phantom{\rule{0ex}{0ex}}T-2\mathit{}mg+2\mathit{}ma=0\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{}\mathit{\left(}i\mathit{\right)}\phantom{\rule{0ex}{0ex}}T\mathit{-}mg\mathit{-}ma\mathit{=}\mathit{0}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{}\mathit{\left(}ii\mathit{\right)}\phantom{\rule{0ex}{0ex}}\mathrm{From}\text{e}\mathrm{quation}\text{s}\mathit{}\mathit{\left(}i\mathit{\right)}\mathit{}\mathrm{and}\mathit{}\mathit{\left(}ii\mathit{\right)}\mathit{,}\phantom{\rule{0ex}{0ex}}T=4\mathit{}ma\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{T}{4m}=\frac{4}{m}\mathit{}\mathrm{m}\mathit{/}{\mathrm{s}}^{\mathit{2}}$

$\mathrm{Now},\mathrm{S}=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)\times \frac{4}{m}\times 1\left[\mathrm{as}u=0\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{2}{m}\right)$

$\mathrm{Net}\mathrm{mass}=2m\mathit{-}m\mathit{=}m\phantom{\rule{0ex}{0ex}}\mathrm{Decrease}\mathrm{in}\mathrm{potential}\mathrm{energy},\phantom{\rule{0ex}{0ex}}\mathrm{P}.\mathrm{E}.=mgh\phantom{\rule{0ex}{0ex}}=m\times g\times \left(\frac{2}{m}\right)\phantom{\rule{0ex}{0ex}}=9.8\times 2=19.6\mathrm{J}$

#### Page No 134:

#### Question 30:

The two blocks in an Atwood machine have masses 2 kg and 3 kg. Find the work done by gravity during the fourth second after the system is released from rest.

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}{m}_{1}=3\mathrm{kg},{m}_{2}=2kg,t=\mathrm{during}{4}^{\mathrm{th}}\mathrm{second}$

From the free-body diagram,

$T-3g+3a=0...\left(i\right)\phantom{\rule{0ex}{0ex}}T-2g-2a=0...\left(ii\right)$

Equating (*i*) and (*ii*), we get:

$3g-3a=2g+2a\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{g}{5}\mathrm{m}/{s}^{2}$

Distance travelled in the 4^{th} second,

$\mathrm{s}\left({4}^{\mathrm{th}}\right)=\frac{a}{2}\left(2n-1\right)\phantom{\rule{0ex}{0ex}}=\frac{\left({\displaystyle \frac{g}{5}}\right)}{2}\left(2\times 4-1\right)\phantom{\rule{0ex}{0ex}}=\frac{7g}{10}=\frac{7\times 9.8}{10}\mathrm{m}$

Net mass $\text{'}m\text{'}={m}_{1}-{m}_{2}=3-2=1\mathrm{kg}$

So, decrease in potential energy,

P.E. = *mgh*

$P.E.=1\times 9.8\times \left(\frac{7}{10}\right)\times 9.8=67.2\mathrm{J}=67\mathrm{J}$

So, work done by gravity during the fourth second = *P.E.*= 67 J

#### Page No 134:

#### Question 31:

Consider the situation shown in the figure (8-E2). The system is released from rest and the block of mass 1kg is found to have a speed 0⋅3 m/s after it has descended a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.

Figure

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}{m}_{1}=4\mathrm{kg},{m}_{2}=1\mathrm{kg},\phantom{\rule{0ex}{0ex}}{v}_{2}=0.3\mathrm{m}/s\phantom{\rule{0ex}{0ex}}{v}_{1}=2\times 0.3=0.6\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\left({v}_{1}=2{v}_{2}\mathrm{in}\mathrm{this}\mathrm{system}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Height}\mathrm{descen}\text{ded}\mathrm{by}\text{the}1\mathrm{kg}\mathrm{block},h=1\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{travelled}\mathrm{by}\text{the}4\mathrm{kg}\mathrm{block},\phantom{\rule{0ex}{0ex}}s=2\times 1=2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Initially}\mathrm{the}\mathrm{system}\mathrm{is}\mathrm{at}\mathrm{rest}.\text{So,u}=0\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{work}\mathrm{energy}\mathrm{theoremwhich}\mathrm{says}\mathrm{that}\phantom{\rule{0ex}{0ex}}\mathrm{change}\mathrm{in}\mathit{}K\mathit{.}E\mathit{.}=\text{W}\mathrm{ork}\mathrm{done}\left(\mathrm{for}\mathrm{the}\mathrm{system}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{1}{2}\right){m}_{1}{\nu}_{1}^{2}+\left(\frac{1}{2}\right){m}_{2}{\mathrm{\nu}}_{2}^{2}=\left(-\mu R\right)s+{m}_{2}gh$

$\frac{1}{2}\times 4\times \left(0.36\right)+\frac{1}{2}\times 1\times \left(0.09\right)[\mathrm{As},R=4g=40\mathrm{N}]\phantom{\rule{0ex}{0ex}}=-\mu 40\times 2+1\times 40\times 1\phantom{\rule{0ex}{0ex}}\Rightarrow 0.72+0.045=-80\mu +10\phantom{\rule{0ex}{0ex}}\Rightarrow \mu =\frac{9.235}{80}=0.12$

So, the coefficient of kinetic friction between the block and the table is 0.12 .

#### Page No 134:

#### Question 32:

A block of mass 100 g is moved with a speed of 5⋅0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=100g=0.1\mathrm{kg},\phantom{\rule{0ex}{0ex}}\mathrm{Velocity}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{at}\mathrm{the}\mathrm{highest}\mathrm{point},\nu =5\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{circular}\mathrm{tube},r=10\mathrm{cm}$

Work done by the block

= Total energy at the highest point − Total energy at the lowest point

$=\left(\frac{1}{2}m{\nu}^{2}+mgh-0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow W=\frac{1}{2}\times \left(0.1\right)\times 25+\left(0.1\right)\times 10\times \left(0.2\right)\phantom{\rule{0ex}{0ex}}\mathrm{As},h=2r=0.2\mathrm{m}\phantom{\rule{0ex}{0ex}}W=1.25+0.2=1.45\mathrm{J}$

So, the work done by the tube on the body is 1.45 joule.

#### Page No 134:

#### Question 33:

A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, calculate the work done against friction (negative of the work done by friction).

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{car},\mathrm{m}=1400\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{h}=10\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{the}\mathrm{car}\mathrm{is}\mathrm{moving}\mathrm{when}\mathrm{the}\mathrm{motor}\mathrm{stops},\mathrm{it}\mathrm{has}\mathrm{kinetic}\mathrm{energy}.\mathrm{Thus}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{i}}=54\mathrm{km}/\mathrm{h}\times \frac{5}{18}=15\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{f}}=0\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{K}=\frac{1}{2}{{\mathrm{mv}}_{\mathrm{f}}}^{2}-\frac{1}{2}{{\mathrm{mv}}_{\mathrm{i}}}^{2}\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{K}=\frac{1}{2}\times 1400\left({0}^{2}-{15}^{2}\right)\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{K}=-157500\mathrm{J}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{gravitational}\mathrm{potential}\mathrm{energy}\mathrm{be}\mathrm{zero}\mathrm{at}\mathrm{the}\mathrm{starting}\mathrm{point}.\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{potential}\mathrm{energy}\mathrm{at}\mathrm{the}\mathrm{terminal}\mathrm{is}\phantom{\rule{0ex}{0ex}}{\mathrm{U}}_{\mathrm{i}}=0\phantom{\rule{0ex}{0ex}}{\mathrm{U}}_{\mathrm{f}}=\mathrm{mgh}\phantom{\rule{0ex}{0ex}}{\mathrm{U}}_{\mathrm{f}}=1400\times 9.8\times 10=137200\mathrm{J}\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{U}={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{U}=137200-0=137200\mathrm{J}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{W}\mathrm{be}\mathrm{the}\mathrm{work}\mathrm{done}\mathrm{against}\mathrm{friction}\mathrm{during}\mathrm{ascent}.\phantom{\rule{0ex}{0ex}}\mathrm{Then}"-\mathrm{W}"\mathrm{is}\mathrm{the}\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{frictional}\mathrm{force}.\phantom{\rule{0ex}{0ex}}-\mathrm{W}=\u25b3\mathrm{K}+\u25b3\mathrm{U}\phantom{\rule{0ex}{0ex}}-\mathrm{W}=-157500+137200\phantom{\rule{0ex}{0ex}}-\mathrm{W}=-20300\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{W}=20300\mathrm{J}$

So, work done against friction is 20,300 joule.

#### Page No 134:

#### Question 34:

A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3⋅2 m high. How much work was required (a) to lift the block from the ground and put it an the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically to the ground (d) it slides down the incline? Take g = 10 m/s^{2}.

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=200g=0.2\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{incline},s=10\mathrm{m},\phantom{\rule{0ex}{0ex}}\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{incline},h=3.2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},g=10\mathrm{m}/{s}^{2}$

(a)

Work done, *W* = *mgh* = 0.2 × 10 × 3.2 =6 .4 J

(b)

Work done to slide the block up the incline

$\mathrm{W}=\left(mg\mathrm{sin}\theta \right)\times \mathrm{s}\phantom{\rule{0ex}{0ex}}=\left(0.2\right)\times 10\times \left(3.2/10\right)\times 10\phantom{\rule{0ex}{0ex}}=6.4\mathrm{J}$

(c)

Let final velocity be *v* when the block falls to the ground vertically.

Change in the kinetic energy = Work done

$\frac{1}{2}m{v}^{2}-0=6.4\mathrm{J}$

$\Rightarrow \nu =8\mathrm{m}/s$

(d)

Let $\nu $ be the final velocity of the block when it reaches the ground by sliding.

$\frac{1}{2}m{\nu}^{2}-0=6.4\mathrm{J}$

$\Rightarrow v=8\mathrm{m}/s$

#### Page No 134:

#### Question 35:

In a children's park, there is a slide which has a total length of 10 m and a height of 8⋅0 m (figure 8-E3). A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. Find (a) the work done by the ladder on the boy as he goes up; (b) the work done by the slide on the boy as he comes down. Neglect any work done by forces inside the body of the boy.

Figure

#### Answer:

Given,

$\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{slide},l=10\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{slide},h=8\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Weight}\mathrm{of}\mathrm{the}\mathrm{boy},mg=200\mathrm{N}$

Friction force,

$F=200\times \left(\frac{3}{10}\right)=60\mathrm{N}$

(a) Work done by the ladder on the boy is zero, as work is done by the boy himself while going up.

(b) Work done against frictional force,

$W\mathit{=}\mu \mathit{}RS\mathit{=}fl=\left(-60\right)\times 10\phantom{\rule{0ex}{0ex}}=-600\mathrm{J}$

#### Page No 134:

#### Question 36:

Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from point *A*, how far away from the track will the particle hit the ground?

Figure

#### Answer:

Given,

Height of the starting point of the track, *H* = 1 m

Height of the ending point of the track, *h* = 0.5 m

Let v be the velocity of the particle at the end point on the track.

Applying the law of conservation of energy at the starting and ending point of the track,we get

$mgH=\frac{1}{2}m{\nu}^{2}+mgh\phantom{\rule{0ex}{0ex}}\Rightarrow g-\left(\frac{1}{2}\right){\nu}^{2}=0.5g\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=2\left(g-0.5g\right)=g\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =\sqrt{g}=3.1\mathrm{m}/\mathrm{s}$

After leaving the track, the body exhibits projectile motion for which,

$\mathrm{\theta}=0\phantom{\rule{0ex}{0ex}}y=-0.5\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{equation}\mathrm{of}\mathrm{motion}\mathrm{along}\mathrm{the}\mathrm{horozontal}\mathrm{direction},\phantom{\rule{0ex}{0ex}}-0.5=\left(u\mathrm{sin}\theta \right)t-\left(\frac{1}{2}\right)g{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.5=4.9\times {t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=0.31\mathrm{sec}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathit{}x=\left(\mathrm{\nu}\mathrm{cos}\mathrm{\theta}\right)t\phantom{\rule{0ex}{0ex}}=3.1\times 0.31=1\mathrm{m}$

So, the particle will hit the ground at a horizontal distance of 1 m from the other end of the track.

#### Page No 134:

#### Question 37:

A block weighing 10 N travels down a smooth curved track *AB* joined to a rough horizontal surface (figure 8-E5). The rough surface has a friction coefficient of 0⋅20 with the block. If the block starts slipping on the track from a point 1⋅0 m above the horizontal surface, how far will it move on the rough surface?

Figure

#### Answer:

Given,

$Weightoftheblock,mg=10\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{Friction}\mathrm{coefficient},\mu =0.2\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{block},H=1\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{velocity}=\mathrm{Final}\mathrm{velocity}=0$

Potential energy of the block at the top of the curved track = Kinetic energy of the block at the bottom of the track

$\Rightarrow K.E.=mgh=10\times 1=10\mathrm{J}$

Again on the horizontal surface the frictional force,

$\mathrm{F}=\mu \mathrm{R}=\mu mg=10\times 1=10\mathrm{J}$

So, the K.E. is used to overcome friction.

$\Rightarrow S=\frac{W}{F}=\frac{10}{2}=5\mathrm{m}$

The block stops after covering 5 m on the rough surface.

#### Page No 134:

#### Question 38:

A uniform chain of mass *m* and length *l* overhangs a table with its two third part on the table. Find the work to be done by a person to put the hanging part back on the table.

#### Answer:

Let '*dx*' be the length of an element at distance *x* from the table.

Mass of the element, '*dm*'$=\left(\frac{m}{l}\right)dx$

Work done to putting back this mass element on the table is

$dW=\left(\frac{m}{l}\right)\times x\times g\times dx$

So, total work done to put $\frac{1}{3}$ part back on the table

$W={\int}_{0}^{1/3}\left(\frac{m}{l}\right)gxdx\phantom{\rule{0ex}{0ex}}\Rightarrow W=\left(\frac{m}{l}\right)g{\left[\frac{{x}^{2}}{2}\right]}^{1/3}\phantom{\rule{0ex}{0ex}}=\frac{mgl}{18l}=\frac{mgl}{18}$

The work to be done by a person to put the hanging part back on the table is $\frac{mgl}{18}$.

#### Page No 134:

#### Question 39:

A uniform chain of length *L* and mass *M* overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is $\mu $. Find the work done by friction during the period the chain slips off the table.

#### Answer:

Let *x* length of the chain be on the table at a particular instant.

Consider a small element of length '*dx'* and mass '*dm*' on the table.

*dm* = $\frac{M}{L}dx$

Work done by the friction on this element is

$dW=\mu \mathit{}Rx=\mu \left(\frac{M}{L}\times gx\right)dx$

Total work done by friction on two third part of the chain,

$W={\int}_{2L/3}^{0}\mu \frac{\mathrm{M}}{L}gxdx\phantom{\rule{0ex}{0ex}}\therefore \mathrm{W}=\mu \frac{M}{L}g{\left[\frac{{x}^{2}}{2}\right]}_{0}^{2L/3}\phantom{\rule{0ex}{0ex}}=-\mu \frac{M}{L}g\left[\frac{4{L}^{2}}{18}\right]\phantom{\rule{0ex}{0ex}}=-\frac{2\mu MgL}{9}$

The total work done by friction during the period the chain slips off the table is $-\frac{2\mu MgL}{9}$.

#### Page No 134:

#### Question 40:

A block of mass 1 kg is placed at point *A* of a rough track shown in figure (8-E6). If slightly pushed towards right, it stops at point *B* of the track. Calculate the work done by the frictional force on the block during its transit from *A* to *B*.

Figure

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=1\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Height}\mathrm{of}\mathrm{point}\mathrm{A},H=1\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Height}\mathrm{of}\mathrm{point}\mathrm{B},h=0.8\mathrm{m}$

Work done by friction = Change in potential energy of the body

$\Rightarrow {\mathrm{W}}_{1}=mgh-mg\mathrm{H}\phantom{\rule{0ex}{0ex}}=1\times 10\left(0.8-1\right)\phantom{\rule{0ex}{0ex}}=-1\times 10\times \left(0.2\right)=-2\mathrm{J}$

The work done by the frictional force on the block during its transit from *A* to *B* is $-$2 joule.

#### Page No 135:

#### Question 41:

A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below, so that it acquires an upward speed of 2⋅ m/s. How high will it rise? Take g = 10 m/s^{2}.

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=5\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Compression}\mathrm{in}\mathrm{the}\mathrm{string}\mathrm{with}\mathrm{the}\mathrm{load},x=10\mathrm{cm}=0.1\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{speed}\mathrm{in}\mathrm{upward}\mathrm{direction},\nu =2\mathrm{m}/s,\phantom{\rule{0ex}{0ex}}h=?,g=10\mathrm{m}/{\mathrm{sec}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{So},F=kx=mg\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{mg}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{50}{0.1}=500\mathrm{N}/\mathrm{m}$

Total energy just after the impulse,

$\mathrm{E}=\frac{1}{2}m{\nu}^{2}+\frac{1}{2}k{x}^{2}...\left(i\right)$

Total energy at a height *h*

$=\frac{1}{2}k{\left(h-x\right)}^{2}-mgh$

On solving, we get:

*h* = 0.2 m

= 20 cm

#### Page No 135:

#### Question 42:

A block of mass 250 g is kept on a vertical spring of spring constant 100 N/m fixed from below. The spring is now compressed 10 cm shorter than its natural length and the system is released from this position. How high does the block rise? Take g = 10 m/s^{2}.

#### Answer:

$\mathrm{Given},\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=250g=0.25\mathrm{kg},\phantom{\rule{0ex}{0ex}}\mathrm{Spring}\mathrm{constant},k=100\mathrm{N}/\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Compression}\mathrm{in}\mathrm{the}\mathrm{string},x=10\mathrm{cm}=0.1\mathrm{m},\phantom{\rule{0ex}{0ex}}\mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},g=10\mathrm{m}/{s}^{2}$

Let the block rises to height *h*.

Applying law of conservation of energy which says that the total energy should always remain conserved.

$\frac{1}{2}k{x}^{2}=mgh\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{1}{2}\left(\frac{k{x}^{2}}{mg}\right)\phantom{\rule{0ex}{0ex}}=\frac{100\times 0.01}{2\times \left(0.250\right)\times 10}\phantom{\rule{0ex}{0ex}}=0.2\mathrm{m}=20\mathrm{cm}$

So, the block rises to 20 cm.

#### Page No 135:

#### Question 43:

Figure (8-E7) shows a spring fixed at the bottom end of an incline of inclination 37°. A small block of mass 2 kg starts slipping down the incline from a point 4⋅8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 m up the incline. Find (a) the friction coefficient between the plane and the block and (b) the spring constant of the spring. Take g = 10 m/s^{2}.

Figure

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=2\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{from}\mathrm{the}\mathrm{spring},{S}_{1}=4.8\mathrm{m},\phantom{\rule{0ex}{0ex}}\mathrm{Comression}\mathrm{in}\mathrm{the}\mathrm{spring},x=20\mathrm{cm}=0.2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Final}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{from}\mathrm{the}\mathrm{spring},{\mathrm{S}}_{2}=1\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{As}\mathrm{\theta}=37\xb0,\phantom{\rule{0ex}{0ex}}\mathrm{sin}37\xb0=0.60=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\mathrm{cos}37\xb0=0.80=\frac{4}{5}$

Applying the work-energy principle for downward motion of the block,

$0-0=mg\mathrm{sin}37\xb0\left(x+4.8\right)-\mu \mathrm{R}\times 5-\frac{1}{2}k{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 20\times \left(0.06\right)\times 5-\mu \times 20\times \left(0.80\right)\times 5-\frac{1}{2}k{\left(0.2\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 60-80\mu -0.02k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 80\mu +0.02k=60...\left(i\right)$

Similarly for the upward motion of the body the equation is

$0-0=\left(-mg\mathrm{sin}37\xb0\right)-\mu \mathrm{R}\times 1+\frac{1}{2}k{\left(-2\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 20\times \left(0.06\right)\times 1-\mu \times 20\times \left(0.80\right)\times 1-\frac{1}{2}k{\left(0.2\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 12-16\mu +0.02k=0$

Adding equations (i) and (ii), we get:

$96\mu =48\phantom{\rule{0ex}{0ex}}\Rightarrow \mu =0.5$

Now putting the value of $\mu $ in equation (i), we get:

*k* = 1000 N/m

#### Page No 135:

#### Question 44:

A block of mass *m* moving at a speed $\nu $ compresses a spring through a distance *x* before its speed is halved. Find the spring constant of the spring.

#### Answer:

Let the velocity of the body at P be $\nu $.

So, the velocity of the body at Q is $\frac{\nu}{2}$.

Energy at point P = Energy at point Q

$\mathrm{So},\frac{1}{2}m{\nu}_{\mathrm{P}}^{2}-\left(\frac{1}{2}\right)m{v}_{\mathrm{Q}}^{2}=\frac{1}{2}k{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}k{x}^{2}=\frac{1}{2}m\left({\mathrm{V}}_{\mathrm{P}}^{2}-{\mathrm{V}}_{\mathrm{Q}}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow k{x}^{2}=m\left({\nu}^{2}-\frac{{\nu}^{2}}{4}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow k{x}^{2}=m\frac{\left(4{\nu}^{2}-{\nu}^{2}\right)}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{3m{v}^{2}}{4{x}^{2}}$

#### Page No 135:

#### Question 45:

Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring.

Figure

#### Answer:

Mass of the body is *m*.

Let the elongation in the spring be* x. *

Applying the law of conservation of energy,

$\frac{1}{2}k{x}^{2}=mgx\phantom{\rule{0ex}{0ex}}\Rightarrow x=2mg\mathit{/}k$

#### Page No 135:

#### Question 46:

,A block of mass *m* is attached to two unstretched springs of spring constants *k*_{1} and *k*_{2} as shown in the figure (8-E9). The block is displaced towards the right through a distance *x* and is released. Find the speed of the block as it passes through the mean position shown.

Figure

#### Answer:

The body is displaced *x* towards the right.

Let *v* be the velocity of the body at its mean position.

Applying the law of conservation of energy,

$\frac{1}{2}m{\nu}^{2}=\frac{1}{2}{k}_{1}{x}^{2}+\frac{1}{2}{k}_{2}{x}^{2}$

$\Rightarrow m{\nu}^{2}={x}^{2}\left({k}_{1}+{k}_{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=\frac{{x}^{2}\left({k}_{1}+{k}_{2}\right)}{m}\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =x\sqrt{\frac{{k}_{1}+{k}_{2}}{m}}$

#### Page No 135:

#### Question 47:

A block of mass *m* sliding on a smooth horizontal surface with a velocity $\overrightarrow{\nu}$ meets a long horizontal spring fixed at one end and with spring constant *k,* as shown in figure (8-E10). Find the maximum compression of the spring. Will the velocity of the block be the same as $\overrightarrow{\nu}$ when it comes back to the original position shown?

Figure

#### Answer:

Let the compression in the spring be* x*.

(a) Applying the law of conservation of energy,

maximum compression in the spring will be produced when the block comes to rest .

so change in kinetic energy of the block due to change in its velocity from u m/s to 0 will be equal to the gain in potential energy of the spring.

change in kinetic energy of the block=$\frac{1}{2}m{v}^{2}-\frac{1}{2}m(0{)}^{2}=\frac{1}{2}m{v}^{2}$

gain in the potential energy of spring=$\frac{1}{2}k{x}^{2}$

$\frac{1}{2}m{\nu}^{2}=\frac{1}{2}k{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=\frac{m{\nu}^{2}}{k}\phantom{\rule{0ex}{0ex}}x=\nu \sqrt{\left(\frac{m}{k}\right)}\phantom{\rule{0ex}{0ex}}$

(b) No. The velocity of the block will not be same when it comes back to the original position. It will be in the opposite direction and the magnitude will be the same if we neglect all losses due friction and spring to be perfectly elastic.

#### Page No 135:

#### Question 48:

A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5 cm (figure 8-E11). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring?

Figure

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=100g=0.1\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Compression}\mathrm{in}\mathrm{the}\mathrm{spring},x=5\mathrm{cm}=0.05\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Spring}\mathrm{constant},k=100\mathrm{N}/\mathrm{m}$

Let *v* be the velocity of the block when it leaves the spring.

Applying the law of conservation of energy,

Elastic potential energy of the spring = Kinetic energy of the block

$\frac{1}{2}m{\nu}^{2}=\frac{1}{2}k{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =x\sqrt{\frac{k}{m}}\phantom{\rule{0ex}{0ex}}=\left(0.05\right)\times \sqrt{\frac{\left(100\right)}{0.1}}\phantom{\rule{0ex}{0ex}}=1.58\mathrm{m}/s$

For the projectile motion,

$\mathrm{\theta}=0\xb0,y=-2\phantom{\rule{0ex}{0ex}}\mathrm{Now},y=\left(u\xb7\mathrm{sin}\mathrm{\theta}\right)t-\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}-2=\left(-\frac{1}{2}\right)\times \left(9.8\right)\times {t}^{2}$

$\Rightarrow t=0.63\mathrm{sec},\phantom{\rule{0ex}{0ex}}\mathrm{So},x=\left(u\mathrm{cos}\mathrm{\theta}\right)t\phantom{\rule{0ex}{0ex}}=\left(1.58\right)\times \left(0.36\right)=1\mathrm{m}$

Therefore, the block hits the ground at 1 m from the free end of the spring in the horizontal direction.

#### Page No 135:

#### Question 49:

A small heavy block is attached to the lower end of a light rod of length *l* which can be rotated about its clamped upper end. What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle?

Figure

#### Answer:

Let the velocity of the body at L is '$\nu $' .

If the body is moving in a vertical plane then we need to find the minimum horizontal velocity which needs to be given to the body (velocity at L).

Also as point H is the highest point in the vertical plane so horizontal velocity at H will be zero.

Applying law of conservation of energy at points L and H,

$\frac{1}{2}m{\nu}^{2}=mgh\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{\nu}^{2}=mg\left(2\mathrm{L}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =\sqrt{\left(4g\mathrm{L}\right)}=2\sqrt{g\mathrm{L}}$

#### Page No 135:

#### Question 50:

Figure (8-E12) shows two blocks *A* and *B*, each of mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block *A* can slide is smooth. Block *A* is attached to a spring of spring constant 40 N/m whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block *A* at the instant it breaks off the surface below it. Take *g* = 10 m/s^{2}.

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{each}\mathrm{block},m=320g=0.32\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Spring}\mathrm{constant},k=40\mathrm{N}/\mathrm{m}\phantom{\rule{0ex}{0ex}}h=40\mathrm{cm}=0.4m\mathrm{and}g=10\mathrm{m}/{s}^{2}$

From the free-body diagram,

$kx\mathrm{cos}\mathrm{\theta}=mg$

As, when the block breaks of the surface below it (i.e. gets dettached from the surface) then *R* =0.

$\Rightarrow \mathrm{cos}\mathrm{\theta}=\frac{mg}{kx}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{0.4}{0.4+x}=\frac{3.2}{40x}\phantom{\rule{0ex}{0ex}}\Rightarrow 16x=3.2x+1.28\phantom{\rule{0ex}{0ex}}\Rightarrow x=0.1\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{So},s=\mathrm{AB}=\sqrt{\left(h+{x}^{2}\right)-{h}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(0.5\right)}^{2}-{\left(0.4\right)}^{2}}=0.3\mathrm{m}$

Let the velocity of body B be $\nu $.

Change in K.E. = Work done (for the system)

$\left(\frac{1}{2}m{u}^{2}+\frac{1}{2}m{\nu}^{2}\right)=-\frac{1}{2}k{x}^{2}+mgs\phantom{\rule{0ex}{0ex}}\Rightarrow \left(0.32\right)\times {\nu}^{2}\phantom{\rule{0ex}{0ex}}=-\left(\frac{1}{2}\right)\times 40\times {\left(1.0\right)}^{2}+\left(0.32\right)\times 10\times \left(0.3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =1.5\mathrm{m}/s$

From the figure,

$\u2206l=h\left(\mathrm{sec}\mathrm{\theta}-1\right)...\left(i\right)$

From the principle of conservation of energy,

$mgs=2\left[\frac{1}{2}m{\nu}^{2}\right]+\frac{1}{2}K\u2206{l}^{2}\phantom{\rule{0ex}{0ex}}mgh\mathrm{tan}\mathrm{\theta}=m{\nu}^{2}+\frac{1}{2}k{h}^{2}{\left(\mathrm{sec}\mathrm{\theta}-1\right)}^{2}...\left(ii\right)$

When the motion of the block breaks of the surface below it (i.e gets dettached from the surface on which it was initially placed) then

$mg=kh\left(\mathrm{sec}\mathrm{\theta}-1\right)\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow 1-\mathrm{cos}\mathrm{\theta}=\frac{\mathit{m}\mathit{g}}{\mathit{k}\mathit{h}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\mathrm{\theta}=1-\frac{\mathit{m}\mathit{g}}{\mathit{k}\mathit{h}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{cos}\mathrm{\theta}=\frac{kh-mg}{kh}\phantom{\rule{0ex}{0ex}}=\frac{40\times 0.4-0.32\times 10}{40\times 0.4}\phantom{\rule{0ex}{0ex}}=0.8$

Putting the value of θ in equation (*ii*), we get:

$0.32\times 10\times 0.4\times 0.75$

$=0.32{\nu}^{2}+\frac{1}{2}40\times {\left(0.4\right)}^{2}{\left(1.25-1\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.96=0.32{\nu}^{2}+0.2\phantom{\rule{0ex}{0ex}}\Rightarrow 0.32{\nu}^{2}=0.72\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =1.5\mathrm{m}/\mathrm{s}$

#### Page No 136:

#### Question 51:

One end of a spring of natural length *h* and spring constant *k* is fixed at the ground and the other is fitted with a smooth ring of mass *m* which is allowed to slide on a horizontal rod fixed at a height *h* (figure 8-E13). Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.

Figure

#### Answer:

$\mathrm{\theta}=37\xb0,l=\mathrm{natural}\mathrm{length}h$

Let the velocity be $\text{'}\nu \text{'}$.

$\mathrm{cos}37\xb0=\frac{\mathrm{BC}}{\mathrm{AC}}=0.8=\frac{4}{5}\phantom{\rule{0ex}{0ex}}{\mathrm{A}}_{\mathrm{C}}=\left(h+x\right)=\frac{5h}{4}$

Applying the law of conservation of energy,

$\frac{1}{2}k{x}^{2}=\frac{1}{2}m{\nu}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =x\sqrt{\left(\frac{k}{m}\right)}=\frac{h}{4}\sqrt{\left(\frac{k}{m}\right)}$

#### Page No 136:

#### Question 52:

Figure (8-E14) shows a light rod of length *l* rigidly attached to a small heavy block at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth *h* below the initial position of the hook and the hook gets into the ring as it reaches there. What should be the minimum value of *h* so that the block moves in a complete circle about the ring?

Figure

#### Answer:

Let v be the minimum velocity required to complete a circle about the ring.

Applying the law of conservation of energy,

Total energy at point A = Total energy at point B

$mgl+\frac{1}{2}m{v}^{2}=mg\left(2l\right)+0\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{2gl}$

Let the rod be released from a height *h*.

Total energy at A = Total energy at B

$mgh=\frac{1}{2}m{\nu}^{2}\phantom{\rule{0ex}{0ex}}mgh=\frac{1}{2}m\left(2gl\right)$

So, *h* = *l*

#### Page No 136:

#### Question 53:

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity $\sqrt{10gl}$, where *l* is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60° with the upward vertical.

#### Answer:

(a) Let the velocity at B be ${v}_{1}$.

$\frac{1}{2}m{\nu}^{2}=\frac{1}{2}m{v}_{1}^{2}+mgl\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}m\left(10gl\right)=\frac{1}{2}m{\nu}_{1}^{2}+mgl\phantom{\rule{0ex}{0ex}}{\nu}_{1}^{2}=8gl$

So, the tension in the string at the horizontal position,

$T=\frac{m{\nu}^{2}}{\mathrm{R}}=m\frac{8gl}{l}\phantom{\rule{0ex}{0ex}}=8mg$

(b) Let the velocity at C be ${v}_{2}$.

$\frac{1}{2}m{\nu}^{2}=\frac{1}{2}m{\nu}_{2}^{2}+mg\left(2l\right)$

$\Rightarrow \frac{1}{2}m10gl=\frac{1}{2}m{\nu}_{2}^{2}+2mgl\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}_{2}^{2}=6gl$

So, the tension in the string is given by

${\mathrm{T}}_{\mathrm{C}}=\frac{m{v}_{2}^{2}}{l}-mg=5mg$

(c) Let the velocity at point D be ${\nu}_{4}$.

Again, $\frac{1}{2}m{\nu}^{2}=\frac{1}{2}m{\nu}_{3}^{2}+mgl\left(1+\mathrm{cos}60\xb0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}_{3}^{2}=7gl$

So, the tension in the string,

${\mathrm{T}}_{\mathrm{D}}=\frac{m{{\nu}_{3}}^{2}}{l}-mg\mathrm{cos}60\xb0\phantom{\rule{0ex}{0ex}}=m\frac{\left(7gl\right)}{l}-0.5mg\phantom{\rule{0ex}{0ex}}=7mg-0.5mg\phantom{\rule{0ex}{0ex}}=6.5mg$

#### Page No 136:

#### Question 54:

A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37° with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.

Figure

#### Answer:

From the figure,

$\mathrm{cos}\mathrm{\theta}=\frac{\mathrm{OC}}{\mathrm{OB}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OC}=\mathrm{O}B\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=\left(0.5\right)\times \left(0.8\right)=0.4\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{CA}=\left(0.5\right)-\left(0.4\right)\phantom{\rule{0ex}{0ex}}=0.1\mathrm{m}$

Total energy at A = Total energy at B

$\frac{1}{2}m{\nu}^{2}=mg\left(\mathrm{A}C\right)\phantom{\rule{0ex}{0ex}}{\nu}^{2}=2\times 10\times \left(0.1\right)=2\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{2}$

So, the tension is given by

$T=\frac{m{\nu}^{2}}{r}+mg\phantom{\rule{0ex}{0ex}}=\left(0.1\right)\left(\frac{2}{0.5}+10\right)=1.4\mathrm{N}$

#### Page No 136:

#### Question 55:

Figure (8-E15) shows a smooth track, a part of which is a circle of radius *R*. A block of mass *m* is pushed against a spring of spring constant *k* fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force *mg* when it reaches the point *P*, where the radius of the track is horizontal.

#### Answer:

Given,

normal force on the track at point P,

N = *mg*

As shown in the figure,

$\frac{m{\nu}^{2}}{\mathrm{R}}=mg\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=g\mathrm{R}...\left(i\right)$

Total energy at point A = Total energy at point P

$\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.\frac{1}{2}k{x}^{2}=\frac{1}{2}m{\nu}^{2}+mg\mathrm{R}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=\frac{mg\mathrm{R}+2mg\mathrm{R}}{k}\phantom{\rule{0ex}{0ex}}[\mathrm{because},{\nu}^{2}=g\mathrm{R}]\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=3mg\mathrm{R}/k\phantom{\rule{0ex}{0ex}}\Rightarrow x=\sqrt{\frac{\left(3mg\mathrm{R}\right)}{k}}$

#### Page No 136:

#### Question 56:

The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of $\sqrt{3gl}$. Find the angle rotated by the string before it becomes slack.

#### Answer:

Suppose the string becomes slack at point P.

Let the bob rise to a height *h*.

*h = l + l *cos θ

From the work-energy theorem,

$\frac{1}{2}m{\nu}^{2}-\frac{1}{2}m{u}^{2}=-mgh\phantom{\rule{0ex}{0ex}}{\nu}^{2}={u}^{2}-2g\left(l+l\mathrm{cos}\theta \right)...\left(i\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again},\frac{m{\nu}^{2}}{l}=mg\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}{\nu}^{2}=lg\mathrm{cos}\mathrm{\theta}...................\left(\mathrm{ii}\right)$

Using equation (*i*) and (*ii*) and the value of u, we get,

$gl\mathrm{cos}\mathrm{\theta}=3gl-2gl-2gl\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}3\mathrm{cos}\mathrm{\theta}=1\phantom{\rule{0ex}{0ex}}\mathrm{\theta}={\mathrm{cos}}^{-1}\left(\frac{1}{3}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left(-\frac{1}{3}\right)$

#### Page No 136:

#### Question 57:

A heavy particle is suspended by a 1⋅5 m long string. It is given a horizontal velocity of $\sqrt{57}\mathrm{m}/\mathrm{s}$. (a) Find the angle made by the string with the upward vertical when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g = 10 m/s^{2}.

#### Answer:

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{string},L=1.5\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{particle},u=\sqrt{57}\mathrm{m}/s\phantom{\rule{0ex}{0ex}}\left(\mathrm{a}\right)mg\mathrm{cos}\mathrm{\theta}=\frac{m{\nu}^{\mathit{2}}}{L}\phantom{\rule{0ex}{0ex}}{\mathrm{\nu}}^{2}=Lg\mathrm{cos}\mathrm{\theta}...\left(\mathrm{i}\right)$

Change in K.E. = Work done

$\frac{1}{2}m{\nu}^{2}-\frac{1}{2}m{u}^{2}=-mgh\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}-57=-2\times 1.5g\left(1+\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=57-3g\left(1+\mathrm{cos}\mathrm{\theta}\right)...\left(ii\right)$

Putting the value of $\nu $ from equation (i),

$15\mathrm{cos}\mathrm{\theta}=57-3g\left(1+\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 15\mathrm{cos}\mathrm{\theta}=57-30-30\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow 45\mathrm{\theta}=27\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\mathrm{\theta}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\theta}={\mathrm{cos}}^{-1}\frac{3}{5}=53\xb0\phantom{\rule{0ex}{0ex}}$

(b) From equation (ii),

$\nu =\sqrt{57-3g\left(1+\mathrm{cos}\mathrm{\theta}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{9}=3\mathrm{m}/\mathrm{s}$

(c) As the string becomes slack at point P, the particle will start executing a projectile motion.

$h=\mathrm{OF}+\mathrm{F}C\phantom{\rule{0ex}{0ex}}=1.5\mathrm{cos}\mathrm{\theta}+\frac{{u}^{2}{\mathrm{sin}}^{2}\mathrm{\theta}}{2g}\phantom{\rule{0ex}{0ex}}=\left(1.5\right)\times \frac{3}{5}+\frac{9\times {\left(0.8\right)}^{2}}{2\times 10}\phantom{\rule{0ex}{0ex}}=1.2\mathrm{m}$

#### Page No 136:

#### Question 58:

A simple pendulum of length *L* with a bob of mass *m* is deflected from its rest position by an angle θ and released (figure 8-E16). The string hits a peg which is fixed at a distance *x* below the point of suspension and the bob starts going in a circle centred at the peg. (a) Assuming that initially the bob has a height less than the peg, show that the maximum height reached by the bob equals its initial height. (b) If the pendulum is released with $\mathrm{\theta}=90\xb0\mathrm{and}x=\mathrm{L}/2$, find the maximum height reached by the bob above its lowest position before the string becomes slack. (c) Find the minimum value of *x*/*L* for which the bob goes in a complete circle about the peg when the pendulum is released from $\mathrm{\theta}=90\xb0$.

Figure

#### Answer:

(i) (ii)

(a) When the bob has an initial height less than the distance of the peg from the suspension point and the bob is released from rest (Fig.(i)),

let body travels from A to B then by the principle of conservation of energy (total energy should always be conserved)

Total energy at A = Total energy at B

$i.e.{\left(K.E.\right)}_{\mathrm{A}}+{\left(P.E.\right)}_{\mathrm{A}}={\left(K.E.\right)}_{\mathrm{B}}+{\left(P.E.\right)}_{\mathrm{B}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(P.E.\right)}_{\mathrm{A}}={\left(P.E.\right)}_{\mathrm{B}}\phantom{\rule{0ex}{0ex}}\mathrm{because}{\left(K.E.\right)}_{\mathrm{A}}={\left(K.E.\right)}_{\mathrm{B}}=0$

So, the maximum height reached by the bob is equal to the initial height of the bob.

(b) When the pendulum is released with θ $=90\xb0\mathrm{and}x=\frac{L}{2},$

Let the string become slack at point C, so the particle will start making a projectile motion.

Applying the law of conservation of emergy

$\left(\frac{1}{2}\right)m{\nu}_{c}^{2}-0=mg\left(\frac{\mathrm{L}}{2}\right)\left(1-\mathrm{cos}\alpha \right)$

[because, distance between A and C in the vertical direction is $\frac{\mathrm{L}}{2}\mathrm{and}$

$\frac{\mathrm{L}}{2}=\left(1-\mathrm{cos}\alpha \right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{V}}_{c}^{2}=g\mathrm{L}\left(1-\mathrm{cos}\alpha \right)...\left(i\right)$

Again, from the free-body diagram (fig. (ii)),

$\frac{m{\nu}_{c}^{2}}{\mathrm{L}/2}=mg\mathrm{cos}\alpha ...\left(ii\right)$

[because, T_{c} = 0]

From equations (i) and (ii),

$g\mathrm{L}\left(1-\mathrm{cos}\alpha \right)=\frac{g\mathrm{L}}{2}\mathrm{cos}\alpha \phantom{\rule{0ex}{0ex}}\Rightarrow 1-\mathrm{cos}\alpha =\frac{1}{2}\mathrm{cos}\alpha \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{2}\mathrm{cos}\alpha =1\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\alpha =\left(\frac{2}{3}\right)...\left(iii\right)$

To find highest position C_{1} upto which the bob can go before the string becomes slack.(as we have found out the value of $\alpha $ so now we want to find the distance of the highest point upto which the bob goes before the string becomes slack,using this value of $\alpha $.

$\mathrm{BF}=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}\times \frac{2}{3}\phantom{\rule{0ex}{0ex}}=\mathrm{L}\left(\frac{1}{2}+\frac{1}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{BF}=\left(\frac{5\mathrm{L}}{6}\right)$

(c) If the particle has to complete a vertical circle at the point C,

$\frac{m{\nu}_{c}^{2}}{\mathrm{L}-x}=mg...\left(i\right)$

Again, applying energy conservation principle between A and C

$\left(\frac{1}{2}\right)m{\nu}_{c}^{2}-0=mg\left(\mathrm{OC}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{1}{2}\right)m{\nu}_{c}^{2}=mg\left\{\mathrm{L}-2\left(\mathrm{L}-\mathrm{x}\right)\right\}\phantom{\rule{0ex}{0ex}}=mg\left(2x-\mathrm{L}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}_{c}^{2}=2g\left(2x-\mathrm{L}\right)...\left(ii\right)$

From equations (i) and (ii),

$g\left(\mathrm{L}-x\right)=2g\left(2x-\mathrm{L}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{L}-x=4x-2\mathrm{L}\phantom{\rule{0ex}{0ex}}\Rightarrow 5x=3\mathrm{L}\phantom{\rule{0ex}{0ex}}\therefore \frac{x}{\mathrm{L}}=\frac{3}{5}=0.6\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\left(\frac{x}{\mathrm{L}}\right)\mathrm{shoule}\mathrm{be}0.6.$

#### Page No 136:

#### Question 59:

A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.

#### Answer:

Let the velocity be $\nu $ when the body leaves the surface.

From the free-body diagram,

$\frac{m{\nu}^{2}}{\mathrm{R}}=mg\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}[\mathrm{normal}\mathrm{reaction}]\phantom{\rule{0ex}{0ex}}{\mathrm{\nu}}^{2}=\mathrm{Rg}\mathrm{cos}\mathrm{\theta}...\left(\mathrm{i}\right)$

Again, from the work-energy principle,

Change in K.E. = Work done

$\Rightarrow \frac{1}{2}m{\nu}^{2}-0=mg\left(\mathrm{R}-\mathrm{R}\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=2g\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)....\left(ii\right)$

From (i) and (ii),

$\mathrm{R}g\mathrm{cos}\mathrm{\theta}=2g\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)$

$3g\mathrm{R}\mathrm{cos}\mathrm{\theta}=2g\mathrm{R}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\mathrm{\theta}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\mathrm{\theta}={\mathrm{cos}}^{-1}\left(\frac{2}{3}\right)$

#### Page No 136:

#### Question 60:

A particle of mass *m* is kept on a fixed, smooth sphere of radius *R* at a position where the radius through the particle makes an angle of 30° with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release? (b) Find the distance travelled by the particle before it loses contact with the sphere.

#### Answer:

(a) When the particle is released from rest, the centrifugal force is zero.

$\mathrm{N}\mathrm{force}=mg\mathrm{cos}\mathrm{\theta}=mg\mathrm{cos}30\xb0\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{2}mg$

(b)

Consider that the particle loses contact with the surface at a point whose angle with the horizontal is $\theta $.

$\mathrm{So},\frac{m{\nu}^{2}}{\mathrm{R}}=mg\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=\mathrm{R}g\mathrm{cos}\mathrm{\theta}...\left(i\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again},\left(\frac{1}{2}\right)m{\nu}^{2}=mg\mathrm{R}\left(\mathrm{cos}30\xb0-\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=2\mathrm{R}g\left(\frac{\sqrt{3}}{2}-\mathrm{cos}\mathrm{\theta}\right)...\left(ii\right)$

From equations (i) and (ii),

$\mathrm{R}g\mathrm{cos}\mathrm{\theta}=2\mathrm{R}g\left[\frac{\sqrt{3}}{2}-\mathrm{cos}\mathrm{\theta}\right]$

$\Rightarrow 3\mathrm{cos}\mathrm{\theta}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\mathrm{\theta}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{\theta}={\mathrm{cos}}^{-1}\frac{1}{\sqrt{3}}$

So, the distance travelled by the particle before losing contact,

$L\mathit{=}R\mathit{}\left(\mathrm{\theta}-\frac{\pi}{6}\right)\phantom{\rule{0ex}{0ex}}\left[\mathrm{because}30\xb0=\left(\frac{\pi}{6}\right)\right]$

Putting the value of θ, we get:

*L* = 0.43 R

#### Page No 136:

#### Question 61:

A particle of mass *m* is kept on the top of a smooth sphere of radius *R*. It is given a sharp impulse which imparts it a horizontal speed $\nu $. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of $v$ for which the particle does not slip on the sphere? (c) Assuming the velocity $v$ to be half the minimum calculated in part, (b) find the angle made by the radius through the particle with the vertical when it leaves the sphere.

Figure

#### Answer:

(a) Radius = R

Horizontal speed = $\nu $

From the above diagram:

Normal force,

$N=mg-\frac{m{v}^{2}}{R}$

(b) When the particle is given maximum velocity, so that the centrifugal force balances the weight, the particle does not slip on the sphere.

$\text{So,}\frac{m{\nu}^{2}}{\mathrm{R}}=mg\phantom{\rule{0ex}{0ex}}\Rightarrow \nu =\sqrt{g\mathrm{R}}$

(c) If the body is given velocity ${\nu}_{1}$ at the top such that,

${\nu}_{1}=\frac{\sqrt{g\mathrm{R}}}{2}\phantom{\rule{0ex}{0ex}}{\nu}_{1}^{2}=\frac{g\mathrm{R}}{4}$

Let the velocity be ${\nu}_{2}$ when it loses contact with the surface, as shown below.

So, $\frac{m{\nu}_{2}^{2}}{\mathrm{R}}=mg\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\nu}}_{2}^{2}=\mathrm{Rg}\mathrm{cos}\mathrm{\theta}...\left(\mathrm{i}\right)$

$\mathrm{Again,}\left(\frac{1}{2}\right)m{\nu}_{2}^{2}-\left(\frac{1}{2}\right)m{\nu}_{1}^{2}\phantom{\rule{0ex}{0ex}}=mg\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}_{2}^{2}={\nu}_{1}^{2}+2g\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)...\left(ii\right)$

From equations (i) and (ii),

$\mathrm{R}g\mathrm{cos}\mathrm{\theta}=\left(\frac{\mathrm{R}g}{4}\right)+2g\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\mathrm{\theta}=\left(\frac{1}{4}+2-2\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 3\mathrm{cos}\mathrm{\theta}=\left(\frac{9}{4}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\theta}={\mathrm{cos}}^{-1}\left(\frac{3}{4}\right)$

#### Page No 137:

#### Question 62:

Figure (8-E17) shows a smooth track which consists of a straight inclined part of length *l* joining smoothly with the circular part. A particle of mass *m* is projected up the incline from its bottom. (a) Find the minimum projection-speed ${\nu}_{0}$ for which the particle reaches the top of the track. (b) Assuming that the projection-speed is $2{\nu}_{0}$ and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is only slightly greater than ${\nu}_{0}$, where will the block lose contact with the track?

#### Answer:

(a) Net force on the particle at A and B,

$\mathrm{F}=mg\mathrm{sin}\mathrm{\theta}$

Work done to reach B from A,

$\mathrm{W}=\mathrm{FS}=mg\mathrm{sin}\mathrm{\theta}l$

Again, work done to reach B to C

$=mgh\phantom{\rule{0ex}{0ex}}=mg\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)$

So, total work done

$=mgl\mathrm{sin}\mathrm{\theta}+mgR\left(1-\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}=mg\left[l\mathrm{sin}\mathrm{\theta}+R\left(1-\mathrm{cos}\mathrm{\theta}\right)\right]$

Now, change in K.E. = Total work done

$\Rightarrow \frac{1}{2}m{\nu}_{2}^{2}=mg\left[l\mathrm{sin}\mathrm{\theta}+\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}_{2}=\sqrt{2g\left[\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)+l\mathrm{sin}\mathrm{\theta}\right]}$

(b) When the block is projected at a speed:

Let the velocity at C be ${\nu}_{0}$.

Applying energy principle,

$\left(\frac{1}{2}\right)m{\nu}_{0}^{2}-\left(\frac{1}{2}\right)m{\left(2{\nu}_{0}\right)}^{2}\phantom{\rule{0ex}{0ex}}=-mg\left[l\mathrm{sin}\mathrm{\theta}+\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{V}}^{2}=4{\nu}_{0}^{2}-2g\left[l\mathrm{sin}g\mathrm{\theta}+\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)\right]\phantom{\rule{0ex}{0ex}}=4.2g\left[l\mathrm{sin}\mathrm{\theta}+\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)\right]-\phantom{\rule{0ex}{0ex}}2g\left[l\mathrm{sin}\mathrm{\theta}+\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)\right]$

So, force acting on the body,

$\mathrm{N}=\frac{{\mathrm{V}}^{2}}{\mathrm{R}}=6mg\left[\left(\frac{l}{\mathrm{R}}\right)\mathrm{sin}\mathrm{\theta}+1-\mathrm{cos}\mathrm{\theta}\right]$

(c) Let the loose contact after making an angle θ.

$\frac{m{\nu}^{2}}{\mathrm{R}}=mg\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=\mathrm{R}g\mathrm{cos}\mathrm{\theta}...\left(i\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again},\frac{1}{2}m{\nu}^{2}=mg\left(\mathrm{R}-\mathrm{R}\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=2g\mathrm{R}\left(1-\mathrm{cos}\mathrm{\theta}\right)...\left(ii\right)$

$\mathrm{From}\left(i\right)\mathrm{and}\left(ii\right),={\mathrm{cos}}^{-1}\left(\frac{2}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\theta}={\mathrm{cos}}^{-1}\left(\frac{2}{3}\right)$

#### Page No 137:

#### Question 63:

A chain of length *l* and mass *m* lies on the surface of a smooth sphere of radius *R* > *l* with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle θ. (c) Find the tangential acceleration $\frac{d\nu}{dt}$ of the chain when the chain starts sliding down.

#### Answer:

Let us consider a small element, which makes angle '*d*θ' at the centre.

$\therefore dm=\rho \left(\frac{m}{\mathrm{L}}\right)\mathrm{R}d\mathrm{\theta}$

(a) Gravitational potential energy of '*dm*' with respect to centre of the sphere

$=\left(dm\right)g\mathrm{R}\mathrm{cos}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=\left(\frac{mg}{\mathrm{L}}\right){\mathrm{R}}^{2}\mathrm{cos}\mathrm{\theta}d\mathrm{\theta}$

$\therefore \mathrm{Total}\mathrm{gravitational}\mathrm{potential}\mathrm{energy},{E}_{\mathrm{P}}=\underset{0}{\overset{\mathrm{L}/\mathrm{R}}{\int}}mg\frac{{\mathrm{R}}^{2}}{\mathrm{L}}\mathrm{cos}\mathrm{\theta}d\mathrm{\theta}\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{P}}=\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\mathrm{sin}\mathrm{\theta}\right]\left[\mathrm{As},\mathrm{\theta}=\frac{\mathrm{L}}{\mathrm{R}}\right]\phantom{\rule{0ex}{0ex}}{E}_{\mathrm{P}}=\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\mathrm{sin}\left(\frac{\mathrm{L}}{\mathrm{R}}\right)$

(b) When the chain is released from rest and slides down through an angle θ,

Change in K.E. of the chain = Change in potential energy of the chain

$=\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\mathrm{sin}\left(\frac{\mathrm{L}}{\mathrm{R}}\right)-\int \frac{g{\mathrm{R}}^{2}}{\mathrm{L}}\mathrm{cos}\mathrm{\theta}d\mathrm{\theta}\phantom{\rule{0ex}{0ex}}=\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\mathrm{sin}\left(\frac{\mathrm{L}}{\mathrm{R}}\right)+\mathrm{sin}\mathrm{\theta}-\mathrm{sin}\left\{\mathrm{\theta}+\left(\frac{\mathrm{L}}{\mathrm{R}}\right)\right\}\right]$

(c) Since,

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\nu}^{2}\phantom{\rule{0ex}{0ex}}=\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\mathrm{sin}\left(\frac{\mathrm{L}}{\mathrm{R}}\right)\right]$

Taking derivative of both sides with respect to '*t*', we get:

$\left(\frac{1}{2}\right)\times 2\nu \times \frac{d\nu}{dt}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\mathrm{cos}\mathrm{\theta}-\frac{d\mathrm{\theta}}{dt}-\mathrm{cos}\left(\mathrm{\theta}+\frac{\mathrm{L}}{\mathrm{R}}\right)\frac{d\mathrm{\theta}}{dt}\right]\phantom{\rule{0ex}{0ex}}\therefore \left[\mathrm{R}-\frac{d\mathrm{\theta}}{dt}\right]\frac{dv}{dt}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{R}}^{2}g}{\mathrm{L}}\times \frac{d\mathrm{\theta}}{dt}\left[\mathrm{cos}\mathrm{\theta}-\mathrm{cos}\left(\mathrm{\theta}+\frac{\mathrm{L}}{\mathrm{R}}\right)\right]\phantom{\rule{0ex}{0ex}}\left(\mathrm{because}\nu =\mathrm{R}\omega =\mathrm{R}\frac{d\mathrm{\theta}}{dt}\right)\phantom{\rule{0ex}{0ex}}\therefore \frac{d\nu}{dt}=\frac{\mathrm{R}g}{\mathrm{L}}\left[\mathrm{cos}\mathrm{\theta}-\mathrm{cos}\left(\mathrm{\theta}+\frac{\mathrm{L}}{\mathrm{R}}\right)\right]$

When the chain starts sliding down,

$\mathrm{\theta}=0$$\xb0$

$\therefore \frac{d\nu}{dt}=\frac{\mathrm{R}g}{\mathrm{L}}\left[1-\mathrm{cos}\left(\frac{\mathrm{L}}{\mathrm{R}}\right)\right]$

#### Page No 137:

#### Question 64:

A smooth sphere of radius *R* is made to translate in a straight line with a constant acceleration *a*. A particle kept on the top of the sphere is released at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.

#### Answer:

Suppose the sphere moves to the left with acceleration '*a*'

Let *m* be the mass of the particle.

The particle '*m*' will also experience inertia due to acceleration '*a*' as it is in the sphere. It will also experience the tangential inertia force $\left[m\left(\frac{d\nu}{dt}\right)\right]$ and centrifugal force $\left(\frac{m{\nu}^{2}}{R}\right)$.

From the diagram,

$m\frac{d\nu}{dt}=ma\mathrm{cos}\mathrm{\theta}+mg\mathrm{sin}\mathrm{\theta}$

$\Rightarrow m\nu \frac{d\nu}{dt}=ma\xb7\mathrm{cos}\mathrm{\theta}\left(\mathrm{R}\frac{d\mathrm{\theta}}{dt}\right)+mg\mathrm{sin}\mathrm{\theta}\left(\mathrm{R}\frac{d\mathrm{\theta}}{dt}\right)\left(\mathrm{because},\nu =\mathrm{R}\frac{d\mathrm{\theta}}{dt}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \nu d\nu =a\mathrm{R}\mathrm{cos}\mathrm{\theta}d\mathrm{\theta}+g\mathrm{R}\mathrm{sin}\mathrm{\theta}d\mathrm{\theta}$

Integrating both sides, we get:

$\frac{{\nu}^{2}}{2}=a\mathrm{R}\mathrm{sin}\mathrm{\theta}-g\mathrm{R}\mathrm{cos}\mathrm{\theta}+\mathrm{C}$

Given: $\mathrm{\theta}=0,\nu =0$

So, $\mathrm{C}=g\mathrm{R}$

$\Rightarrow \frac{{\nu}^{2}}{2}=a\mathrm{R}\mathrm{sin}\mathrm{\theta}-g\mathrm{R}\mathrm{cos}\mathrm{\theta}+g\mathrm{R}\phantom{\rule{0ex}{0ex}}\Rightarrow {\nu}^{2}=2\mathrm{R}\left(a\mathrm{sin}\mathrm{\theta}+g-g\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \nu ={\left[2\mathrm{R}\left(a\mathrm{sin}\mathrm{\theta}+g-g\mathrm{cos}\mathrm{\theta}\right)\right]}^{1/2}$

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