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#### Page No 130:

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:

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:

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:

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:

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:

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:

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:

(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:

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:

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:

(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:

(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:

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:

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:

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:

(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:

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

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

(b) Work done by the kinetic friction = final kinetic energy $-$ initial kinetic energy
= $\frac{1}{2}m\left({v}_{0}{\right)}^{2}-\frac{1}{2}m\left(v-{v}_{0}{\right)}^{2}$
= $-\frac{1}{2}m{v}^{2}+mv{v}_{0}$

#### Page No 131:

(d) the speed does not depend on the initial direction

As the stone 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

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

From the above expression, we can say that the maximum speed with which stone hits the ground does not depend on the initial direction.

#### Page No 131:

(b) 2E

Let xA and xB be the extensions produced in springs A and B, respectively.
Restoring force on spring A, $F={k}_{\mathrm{A}}{x}_{\mathrm{A}}$   ...(i)
Restoring force on spring B, $F={k}_{\mathrm{B}}{x}_{\mathrm{B}}$   ...(ii)

From (i) and (ii), we get:
${k}_{\mathrm{A}}{x}_{\mathrm{A}}={k}_{\mathrm{B}}{x}_{\mathrm{B}}$

It is given that kA = 2kB
$\therefore {x}_{\mathrm{B}}=2{x}_{\mathrm{A}}$

Energy stored in spring A:
$E=\frac{1}{2}{k}_{\mathrm{A}}{{x}_{\mathrm{A}}}^{2}$   ...(iii)

Energy stored in spring B:

#### Page No 131:

(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:

(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.

#### Page No 131:

(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:

(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:

(c) Potential energy

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

#### Page No 131:

(c) mgvt sin2θ

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

#### Page No 131:

(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:

(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

#### Page No 131:

(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

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

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:

(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 132:

(c) its kinetic energy is constant
(d) it moves in a circular path.

When the force on a particle is always perpendicular to its velocity, the work done by the force on the particle is zero, as the angle between the force and velocity is 90$°$. So, kinetic energy of the particle will remain constant. The force acting perpendicular to the velocity of the particle provides centripetal acceleration that causes the particle to move in a circular path.

#### Page No 132:

(d) acceleration of the block

Acceleration of the block will be the same to both the observers. The respective kinetic energies of the observers are different, because the block appears to be moving with different velocities to both the observers. Work done by the friction and the total work done on the block are also different to the observers.

#### Page No 132:

(a) the path taken by the suitcase
(b) the time taken by you in doing so

Work done by us on the suitcase is equal to the change in potential energy of the suitcase.
i.e., W = mgh
Here, mg is the weight of the suitcase and h is height of the table.

Hence, work done by the conservative (gravitational) force does not depend on the path.

#### Page No 132:

(a) the force is always perpendicular to its velocity
(c) the object is stationary but the point of application of the force moves on the object
(d) the object moves in such a way that the point of application of the force remains fixed.

No work is done by a force on an object if the force is always perpendicular to its velocity. Acceleration does not always provide the direction of motion, so we cannot say that no work is done by a force on an object if it is always perpendicular to the acceleration. Work done is zero when the displacement is zero.
In a circular motion, force provides the centripetal acceleration. The angle between this force and the displacement is 90$°$, so work done by the force on an object is zero.

#### Page No 132:

(a) The string becomes slack when the particle reaches its highest point.
(d) The particle again passes through the initial position.

The string becomes slack when the particle reaches its highest point. This is because at the highest point, the tension in the string is minimum. At this point, potential energy of the particle is maximum, while its kinetic energy is minimum. From the law of conservation of energy, we can see that the particle again passes through the initial position where its potential energy is minimum and its kinetic energy is maximum.

#### Page No 132:

(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:

(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:

(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
$⇒$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:

Total mass of the system (cyclist and bike),

Initial velocity of the system,

Final velocity of the system,

From work-energy theorem, we have:

#### Page No 132:

Resisting force acting on the box,
Displacement of the box, S = 4 m
Also,

∴ Work done by the resisting force,

#### Page No 132:

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.

#### Page No 132:

Given:

Work done by the force,

Acceleration of the particle is,

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

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

#### Page No 132:

Initial position vector,
$\stackrel{\to }{{r}_{1}}=2\stackrel{\to }{i}+3\stackrel{\to }{j}$
Final position vector,
${\stackrel{\to }{r}}_{2}=3\stackrel{\to }{i}+2\stackrel{\to }{j}$
So, displacement vector,

#### Page No 132:

Given:

So, force applied by the man on the box,

#### Page No 133:

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,

#### Page No 133:

Here, R is the normal reaction of the block.

As the block is moving with uniform speed,
$f=mg\mathrm{sin}37°$

So, work done against the force of friction,

#### Page No 133:

(a)

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

For the block of mass m,

(b)
Frictional force acting on the smaller block,

(c) Work done, w = f1s [where s = d]

#### Page No 133:

Given:

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

(a) From the figure,

From (i) and (ii),

So, work done by the person,

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

Putting the value in equation (iv),

#### Page No 133:

Given:

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

∴ Work done when the force is in horizontal direction,

#### Page No 133:

Retardation of the car,

#### Page No 133:

Force needed to accelerate the car,
$\mathrm{F}=ma=500×8=400\mathrm{N}$

Given,

#### Page No 133:

(a)
Given:

From the above figure,

(b) If

So, work done

(c)

#### Page No 133:

Given:

(a) t = 1 sec

So,

Work done by the applied force,

(b)
So, work done by weight,

So, frictional force,

Work done by the friction forces,

#### Page No 133:

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

Again, work done against friction,

#### Page No 133:

Given:

Power of a lamp,

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

So, power in watt $\left(\mathrm{J}/\mathrm{sec}\right)=\left(\frac{441×{10}^{5}}{60×60}\right)$
Therefore, the number of 100 W lamps that can be lit using this energy,
$n=\frac{441×{10}^{5}}{3600×100}=122.5\approx 122$

Given:

#### Page No 133:

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,

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

#### Page No 133:

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

#### Page No 133:

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)

(b)

(c)
To maintain uniform speed, she had to exert 4900 J of energy to overcome friction.
Power exerted by her to overcome frcition,

#### Page No 133:

Given:
Height through which water is lifted, h = 10 m

Power delivered by the engine,

1 hp = 746 w

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

#### Page No 133:

1 hp = 764 watt

Horsepower used by demonstrator

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

Therefore, power used by the demonstrator to lift and throw the stone is 5.14$×$10-3 hp.

#### Page No 133:

Given:
Mass of the metal,
Distance, s = 12 m
Time taken, t = 1 minute = 60 s
Force applied by the engine to lift the metal,
F = mg

#### Page No 133:

The specifications given by the company are:

So, the maximum acceleration that can be produced,

So, the driving force,

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:

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

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

#### Page No 134:

From the free-body diagram,

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

Distance travelled in the 4th second,

Net mass
So, decrease in potential energy,
P.E. = mgh

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

#### Page No 134:

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

#### Page No 134:

Work done by the block
= Total energy at the highest point − Total energy at the lowest point

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

#### Page No 134:

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

#### Page No 134:

(a)
Work done, W = mgh = 0.2 × 10 × 3.2 =6 .4 J

(b)
Work done to slide the block up the incline

(c)
Let final velocity be v when the block falls to the ground vertically.
Change in the kinetic energy = Work done

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

#### Page No 134:

Given,

Friction force,

(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,

#### Page No 134:

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

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

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

#### Page No 134:

Given,

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

Again on the horizontal surface the frictional force,

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

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

#### Page No 134:

Let 'dx' be the length of an element at distance x from the table.
Mass of the element, 'dm'
Work done to putting back this mass element on the table is

$dW=\left(\frac{m}{l}\right)×x×g×dx$

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

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

#### Page No 134:

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

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

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

#### Page No 134:

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

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

#### Page No 135:

Total energy just after the impulse,

Total energy at a height h

On solving, we get:
h = 0.2 m
= 20 cm

#### Page No 135:

Let the block rises to height h.
Applying law of  conservation of energy which says that the total energy should always remain conserved.

So, the block rises to 20 cm.

#### Page No 135:

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

Similarly for the upward motion of the body the equation is

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

Now putting the value of $\mu$ in equation (i), we get:
k = 1000 N/m

#### Page No 135:

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

#### Page No 135:

Mass of the body is m.
Let the elongation in the spring be x.

Applying the law of conservation of energy,

#### Page No 135:

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,

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

#### Page No 135:

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=$\frac{1}{2}m{v}^{2}-\frac{1}{2}m\left(0{\right)}^{2}=\frac{1}{2}m{v}^{2}$
gain in the potential energy of spring=$\frac{1}{2}k{x}^{2}$

(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:

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

For the projectile motion,

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

#### Page No 135:

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,

#### Page No 135:

From the free-body diagram,

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

Let the velocity of body B be $\nu$.
Change in K.E. = Work done (for the system)

From the figure,

From the principle of  conservation of energy,

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

Putting the value of θ in equation (ii), we get:
$0.32×10×0.4×0.75$

#### Page No 136:

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

Applying the law of conservation of energy,

#### Page No 136:

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

Let the rod be released from a height h.

Total energy at A = Total energy at B

So, h = l

#### Page No 136:

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

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

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

So, the tension in the string is given by

(c) Let the velocity at point D be ${\nu }_{4}$.
Again,
So, the tension in the string,

#### Page No 136:

From the figure,

Total energy at A = Total energy at B

So, the tension is given by

#### Page No 136:

Given,
normal force on the track at point P,
N = mg
As shown in the figure,

Total energy at point A = Total energy at point P

#### Page No 136:

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,

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

#### Page No 136:

Change in K.E. = Work done

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

(b) From equation (ii),

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

#### Page No 136:

(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

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

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

Applying the law of conservation of emergy

[because, distance between A and C in the vertical direction is

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

[because, Tc = 0]
From equations (i) and (ii),

To find highest position C1 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 $\alpha$.

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

Again, applying energy conservation principle between A and C

From equations (i) and (ii),

#### Page No 136:

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

From the free-body diagram,

Again, from the work-energy principle,
Change in K.E. = Work done

From (i) and (ii),

#### Page No 136:

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

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

From equations (i) and (ii),

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

Putting the value of θ, we get:
L = 0.43 R

#### Page No 136:

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.

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

(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,

From equations (i) and (ii),

#### Page No 137:

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

Work done to reach B from A,

Again, work done to reach B to C

So, total work done

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

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

Let the velocity at C be ${\nu }_{0}$.
Applying energy principle,

So, force acting on the body,

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

#### Page No 137:

Let us consider a small element, which makes angle 'dθ' at the centre.

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

(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

(c) Since,

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

When the chain starts sliding down,
$\mathrm{\theta }=0$$°$

#### Page No 137:

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,

Integrating both sides, we get:

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

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