Hc Verma I Solutions for Class 11 Science Physics Chapter 6 Friction are provided here with simple step-by-step explanations. These solutions for Friction are extremely popular among Class 11 Science students for Physics Friction 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 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma I Solutions. All Hc Verma I Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 95:

#### Answer:

It is not necessary that the friction coefficient is always less than 1. When the friction is stronger than the normal reaction force, the coefficient of friction is greater than 1. For example, silicon rubber has the coefficient of friction greater than 1.

#### Page No 95:

#### Answer:

It is easier to push a heavy block from behind than from the top because when we try to push a heavy block from the top, we increase the normal reaction force, which, in turn, increases the friction between the object and the ground (see the figure).

#### Page No 95:

#### Answer:

The person started with zero initial velocity, covered a 1 km distance and ended with zero velocity, so the acceleration is zero. Hence, the average friction force is zero.

#### Page No 95:

#### Answer:

It is difficult to walk on solid ice because the coefficient of friction between our foot and ice is very less; hence, a person trying to walk on solid ice may slip.

#### Page No 95:

#### Answer:

No, we cannot accelerate or stop a car on a frictionless horizontal road. The car will not move on a frictionless surface because rolling is not possible without friction.

#### Page No 95:

#### Answer:

In the first case, the normal reaction force is equal to the weight of the stone, hence the stone slides easily because the friction force is very less. However, in the second case, a small piece of wood is sandwiched, which increases the normal reaction force on the wood due to the weight of the door. Hence, greater the normal reaction force on the wood, the greater will be the frictional force between wood and the floor.

#### Page No 96:

#### Answer:

(a) The coefficient of friction between the card and the coin should be small.

(b) The coin should be heavy.

(c) If the card is pushed gently, the experiment fails because the frictional force gets more to time to act and it may gain some velocity and move with the card.

#### Page No 96:

#### Answer:

No, a tug of war cannot be won on a frictionless surface because the tension in the rope on both the sides of both the teams will be same. So, to win, one of the teams must apply some greater force, which is the force of friction.

#### Page No 96:

#### Answer:

The normal reaction force on a level road is *mg,* whereas on an inclined plane it is *mg *cos θ, which means that on an incline road the friction force between the tyre and the road is less. Hence, tyres have less grip on an incline plane and better grip on a level road.

#### Page No 96:

#### Answer:

By throwing the bag in one direction, we gain some velocity in the opposite direction as per the law of conservation of linear momentum. In this way we can come out of the ice easily.

#### Page No 96:

#### Answer:

The coefficient of friction increases between two highly smooth surfaces because the atoms of both the materials come very closer to each and the number of bonds between them increase.

#### Page No 96:

#### Answer:

(b) decrease

According to the first law of limiting friction,

*f* = *μN*

where *f* is the frictional force

*N* is the normal reaction force

*μ* is the coefficient of static friction

and

*N* = *mg *－ *F*cosθ

where *m* is the mass of the body

*F* is the contact force acting on the body

If we decrease the angle between this contact force and the vertical, then *F*cosθ increases and the normal reaction force (*N*) as well as the frictional force (*f*) decrease.

#### Page No 96:

#### Answer:

(b) smaller friction

According to the first law of the limiting friction,

*f* = *μN*

where *f* is the frictional force

*μ* is the coefficient of friction

*N *is the normal reaction force

When we take smaller steps on ice, the normal reaction force exerted by the ice is small. Therefore, the smaller steps ensure smaller friction.

#### Page No 96:

#### Answer:

(c) *Mg* ≤** ***F* ≤ *Mg* $\sqrt{1+{\mathrm{\mu}}^{2}}$

Let *T* be the force applied on an object of mass *M*.

If *T *= 0, *F*_{min} = *Mg.*

If *T* is acting in the horizontal direction, then the body is not moving.

∴ $T=\mathrm{\mu}\left(mg\right)$

${F}_{\mathrm{max}}=\sqrt{(Mg{)}^{2}+(T{)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{(Mg{)}^{2}+(\mathrm{\mu}Mg{)}^{2}}$

Thus, we have:

$Mg\le F\le Mg\sqrt{1+(\mathrm{\mu}{)}^{2}}$

#### Page No 96:

#### Answer:

d) >500 N for time ∆*t*_{1} and ∆*t*_{3} and 500 N for ∆*t*_{2}_{.}

During the time interval ∆*t*_{2}, the scooter is moving with a constant velocity, which implies that the force exerted by the seat on the man is 500 N (for balancing the weight of the man).

During the time interval ∆*t*_{1} and ∆*t*_{3},_{ }the scooter is moving with constant acceleration and deceleration, which implies that a frictional force is also applied. Therefore, the net force exerted by the seat on the man should be >500 N.

#### Page No 96:

#### Answer:

(d) the system cannot remain in equilibrium.

Since the wall is smooth and the surface of *A* and *B *in contact are rough, the net vertical force on the system is in the downward direction. Hence, the system cannot remain in equilibrium.

#### Page No 96:

#### Answer:

(a) is upward

The normal reaction force on the system (comprising of wall and contact surface of *A* and *B*)* *is provided by *F. *As can be seen from the figure, the weight of *A* and *B* is in the downward direction. Therefore, the frictional force *f*_{A} and *f*_{BA} (friction on B due to A) is in upward direction.

#### Page No 96:

#### Answer:

(c) is same for both cars

Given: both the cars have same initial speed.

Let the masses of the two cars be *m*_{1} and *m*_{2}_{.}

Frictional force on car with mass *m*_{1} = μ*m*_{1} g

So, the deceleration due to frictional force = $\frac{\mathrm{\mu}{m}_{1}\mathrm{g}}{{m}_{1}}=\mathrm{\mu g}$

Frictional force on car with mass *m*_{2} = μ*m*_{2} g

So, the deceleration due to frictional force = $\frac{\mathrm{\mu}{m}_{2}\mathrm{g}}{{m}_{2}}=\mathrm{\mu g}$

As both the acceleration are same, from the second equation of motion

$s=ut+\frac{1}{2}a{t}^{2}$

Thus, we can say that both the cars have same minimum stopping distance.

#### Page No 96:

#### Answer:

(b) apply the brakes hard enough to just prevent slipping

When we apply hard brakes just enough to prevent slipping on wheels, it provides optimum normal reaction force, which gives the maximum friction force between tyres of the car and the road.

#### Page No 96:

#### Answer:

(d) all of three are possible.

We know that

*N *= *mg *cos θ˚

*f*_{max} = μ*N** = *μ*mg** *cos θ

where *N* = normal reaction force

*f*_{max} = frictional force

θ = angle of inclination

μ = coefficient of friction

When the block just begins to slide, it means

*mg *sin θ = *f*_{max}

*mg *sin θ_{ = }μ*mg** *cos θ

μ = tan θ

and the coefficient of friction depends on the angle of inclination (θ) and does not depend on mass.

Now consider the block sliding condition:

*mg *sin θ − *f*_{max} = *ma*

*mg *sin θ_{ }−_{ }μ*mg** *cos θ* = ma*

∴* a = g*(sin θ* *− μ cos θ)

From the above equation it is clear that acceleration does not depend on the mass but depends on θ and μ.

#### Page No 96:

#### Answer:

(a) μ < μ', *M* < *M*'

Let *T* be the force applied by the boy on the block.

Free body diagram for the box:

The condition for preventing the slide is

* **f*_{max} > *T*

*μ'M'g *>* T * (i)

Now see the free body diagram of a boy of mass *M*:

*f*_{max} = μ*mg*

The condition for preventing the slide is

*f*_{max} >* T*

μ*mg* > *T*

The condition for sliding the entire system (block and boy) is

*f'* > *f* (block is not slide)

μ'*M'g* > μ*mg*

μ'*M**'* > μ*m*

μ < μ'

*m* < *M'*

#### Page No 97:

#### Answer:

(a) *F *> *F*_{N}

(b) *F* > *f*

(d) *F*_{N} − *f* < *F* < *F*_{N} + *f*

The system is in equilibrium condition when *F* = *f.*

Hence, the net horizontal force is zero.

*f* = μ*F*_{N}

*F > **F*_{N}

*f* = *F*_{N}_{ }and 0 ≤ μ ≤ 1

Therefore, we can say that *F *> *f.* So the net horizontal force is nonzero.

*F* > *f*, and so the net horizontal force is zero.

*F*_{N}* > **f$\Rightarrow $F*_{N} > μ*F*_{N}*$\Rightarrow $*μ < 1

Here, the given relation between *F* and *f* i.e

*F* > *f *and *f* = μ*F*_{N}_{ } will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.

*F*_{N} − *f* < *F* < *F*_{N} + *f*

∵ *f* = μ*F*_{N}

$\frac{f}{\mu}-f<F<\frac{f}{\mu}+f\phantom{\rule{0ex}{0ex}}f\left(\frac{1-\mu}{\mu}\right)<F<f\left(\frac{1+\mu}{\mu}\right)$

For the above relation, we can say that *F ≠ f *and so the net horizontal force is nonzero.

#### Page No 97:

#### Answer:

(b) the force of friction between the bodies is zero

(d) the bodies may be rough but they don't slip on each other

The contact force exerted by a body A on another body B is equal to the normal force between the bodies. Therefore, we can conclude that the force of friction between the bodies is zero or the bodies may be rough but they don't slip on each other.

#### Page No 97:

#### Answer:

(b) Coefficient of static friction is always greater than the coefficient of kinetic friction.

(c) Limiting friction is always greater than the kinetic friction.

(d) Limiting friction is never less than the static friction.

All the above statements are correct. The static friction is sometimes less than the kinetic friction.

#### Page No 97:

#### Answer:

(c) The graph is a straight line of slope 45° for small *F* and a straight line parallel to the *F*-axis for large *F*.

(d) There is a small kink on the graph.

When force *F *is applied* *on the block, the force of friction *f *comes into play. As we increase the applied *F, * the static friction force adjusts itself to become (equal) to the applied force *F* and goes upto its maximum value equal to limiting friction force.After this ,it is treated as a constant force (i.e . now its value does not change until and unless the body starts moving). If the applied force *F* is greater than the limiting friction force, then the kinetic friction force comes into play at that time. The kinetic friction force is always less than the limiting friction force.

#### Page No 97:

#### Answer:

(a) is towards east if the vehicle is accelerating

(b) is zero if the vehicle is moving with a uniform velocity

When the vehicle is accelerating, the force is applied (by the tyre on the road) in west direction .That causes a net resultant frictional force acting in east direction. Due to this force of friction only ,the car is moving in east direction.

When the vehicle is moving with a uniform velocity, the force of friction on the wheels of the vehicle by the road is zero.

#### Page No 97:

#### Answer:

Let *m* be the mass of the body.

From the free body diagram,

*R* − *mg *= 0

(where *R* is the normal reaction force and *g *is the acceleration due to gravity)

⇒ *R *= *mg * (1)

Again *ma* − *μ _{k}R* = 0

(where

*μ*is the coefficient of kinetic friction and

_{k}*a*is deceleration)

or

*ma*=

*μ*

_{k}RFrom Equation (1),

*ma*=

*μ*

_{k}mg⇒

*a*=

*μ*

_{k}g⇒ 4 =

*μ*

_{k}g$\Rightarrow {\mu}_{\mathrm{k}}=\frac{4}{g}=\frac{4}{10}=0.4$

Hence, the coefficient of the kinetic friction between the block and the plane is 0.4.

#### Page No 97:

#### Answer:

Friction force acting on the block will decelerate it.

Let the deceleration be '*a*'.

Using free body diagram

*R* − *mg* = 0

(where *R *is the normal reaction force)

⇒ *R* = *mg* (1)

Again, *ma* − μ_{k}*R* = 0

(where μ_{k }is the coefficient of kinetic friction)

From Equation (1),

⇒ *ma* = μ_{k}*mg*

⇒ *a* = μ_{k}*g* = 0.1 × 10

= 1 m/s^{2}

Given:

initial velocity, *u* = 10 m/s

final velocity, *v* = 0 m/s (block comes to rest)

*a* = −1 m/s^{2} (deceleration)

Using equation of motion *v*^{2} $-$ *u*^{2} = 2*as*

(where *s* is the distance travelled before coming to* *rest)

$s=\frac{{v}^{2}-{u}^{2}}{2a}$

On substituting the respective values, we get

$=0-\frac{{10}^{2}}{2}\left(-1\right)\phantom{\rule{0ex}{0ex}}=\frac{100}{2}=50\mathrm{m}$

Therefore, the block will travel 50 m before coming to rest.

#### Page No 97:

#### Answer:

A block of mass *m* is kept on a horizontal table. If force is applied on the block, a friction force will be there: *p* → frictional force and *F* → applied force

So, friction force is equal to the applied force. One of the case is that the friction force is equals to zero when the applied force is equal to zero.

#### Page No 97:

#### Answer:

Free body diagram for the block is as follows:

From the above diagram:

*R* − *mg* cos θ = 0

⇒ *R* = *mg* cos θ (1)

For the block, *u* = 0 m/s, *s* = 8 m and *t* = 2 s.

According to the equation of motion

$s=ut+\frac{1}{2}a{t}^{2}$

$s=0+\frac{1}{2}a{2}^{2}\phantom{\rule{0ex}{0ex}}8=2a$

*a* = 4 m/s^{2}

Again,

μ_{k}*R* + *ma* − *mg* sin θ = 0

(where μ_{k }is the coefficient of kinetic friction)

From Equation (1):

μ_{k}*mg* cos θ + *ma* − *mg* sin θ = 0

⇒ *m* (μ_{k}*g* cos θ + *a* − *g* sin θ) = 0

⇒ μ_{k} × 10 × cos 30° = *g* sin 30° − *a*

$\Rightarrow {\mathrm{\mu}}_{\mathrm{k}}\times 10\frac{\sqrt{3}}{2}=10\times \left(\frac{1}{2}\right)-4\phantom{\rule{0ex}{0ex}}\Rightarrow \left(5\sqrt{3}\right){\mathrm{\mu}}_{\mathrm{k}}=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\mu}}_{\mathrm{k}}=\left(\frac{1}{5\sqrt{3}}\right)=0.11$

Therefore, the coefficient of kinetic friction between the block and the surface is 0.11.

#### Page No 97:

#### Answer:

Free body diagram of the block for this case is as follows:

From the adove diagram:

*F *− *ma* − μ_{k}*R* + *mg* sin 30° = 0

4 − 4*a* − μ_{k}*R* + 4*g* sin 30° = 0 (1)

*R* − 4*g* cos 30° = 0 (2)

⇒ *R* = 4*g* cos 30° = 0

Substituting the values of *R* in Equation (1) we get

4 − 4*a** *− 0.11 × 4*g* cos 30° + 4*g* sin 30° = 0

$\Rightarrow 4-4a-0.11\times 4\times 10\times \frac{\sqrt{3}}{2}+4\times 10\times \frac{1}{2}=0$

4 − 4*a* − 3.81 + 20 = 0

4 − 4*a* − 3.18 + 20 = 0

*a* ≈ 5 m/s^{2}

For the block, *u* = 0, *t* = 2 s and *a* = 5 m/s^{2}.

According to the equation of motion,

$s=ut+\frac{1}{2}a{t}^{2}$

$=0+\left(\frac{1}{2}\right)5\times {2}^{2}\phantom{\rule{0ex}{0ex}}=10\mathrm{m}$

Therefore, the block will move 10 m.

#### Page No 97:

#### Answer:

(a) To make the block move up the incline, the applied force should be equal and opposite to the net force acting down the incline.

Applied force = μ*R* + 2*g* sin 30° (1)

(where μ is the coefficient of static friction)

*R* = *mg *cos 30°

Substituting the respective values in Equation (1), we get

$=0.2\times \left(9.8\right)\sqrt{3}+2\times 9.8\times \left(\frac{1}{2}\right)$

3.39 + 9.8 $\approx $ 13 N

With this minimum force, the body moves up the incline with a constant velocity as the net force on it is zero.

(b) Net force acting down the incline is given by

*F *= 2*g* sin 30° − μ*R*

$=2\times 9.8\times \frac{1}{2}-3.99\phantom{\rule{0ex}{0ex}}=6.41\mathrm{N}$

Because *F* = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.

#### Page No 97:

#### Answer:

Using the free body diagram,

*g* = 10 m/s^{2}, *m* = 2 kg, θ = 30 and μ = 0.2

*R* − *mg* cos θ − *F* sin θ = 0

⇒ *R* = *mg* cos θ + *F* sin θ (1)

and

*mg *sin θ + μ*R* − *F* cos θ = 0

⇒ *mg* sin θ + μ(*mg* cos θ + *F* sin θ) − *F* cos θ = 0

⇒ *mg* sin θ + μ*mg* cos θ + μ*F* sin θ − *F* cos θ = 0

$\Rightarrow F=\frac{\left(mg\mathrm{sin}\mathrm{\theta}+\mathrm{\mu}mg\mathrm{cos}\mathrm{\theta}\right)}{\left(\mathrm{\mu}\mathrm{sin}\mathrm{\theta}-\mathrm{cos}\mathrm{\theta}\right)}(\mathrm{\theta}=30\xb0)\phantom{\rule{0ex}{0ex}}=\frac{\left(2\times 10\times \left({\displaystyle \frac{1}{2}}\right)+0.2\times 2\times 10\times {\displaystyle \frac{\sqrt{3}}{2}}\right)}{0.2\times \left({\displaystyle \frac{1}{2}}\right)-\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{13.464}{0.76}=17.7\mathrm{N}=17.5\mathrm{N}$

Therefore, while pushing the block to move up on the incline, the required force is 17.5 N.

#### Page No 97:

#### Answer:

Let *m* be the mass of the boy.

From the above diagram:

*R* − *mg* cos 45° = 0

$R=mg\mathrm{cos}45\xb0=\frac{mg}{\sqrt{2}}\left(1\right)$

Net force acting on the boy, making him slide down

= *mg* sin 45° − μ*R*

= *mg* sin 45° − μ*mg* cos 45°

$=m\times 10\times \left(\frac{1}{\sqrt{2}}\right)-0.6\times m\times 10\times \left(\frac{1}{\sqrt{2}}\right)\phantom{\rule{0ex}{0ex}}=m\left(5\sqrt{2}-3\sqrt{2}\right)=m\times 2\times \sqrt{2}$

The acceleration of the boy $=\frac{\mathrm{Force}}{\mathrm{Mass}}$

$=\frac{m\left(2\sqrt{2}\right)}{m}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}\mathrm{m}/{\mathrm{s}}^{2}$

#### Page No 97:

#### Answer:

Let *a* be the acceleration of the body sliding down.

From the above diagram:

*R* − *mg* cos θ = 0

⇒ *R* = *mg* cos θ (1)

and

*ma* + *mg* sin θ − μ*R* = 0

⇒ $a=\frac{mg(\mathrm{sin}\theta -\mathrm{\mu}cos\theta \mathit{)}}{m}=g(\mathrm{sin}\theta -\mathrm{\mu}cos\theta \mathit{)}$

For the first half metre, *u* = 0, *s* = 0.5 m and *t* = 0.5 s.

According to the equation of motion,

* v* = *u* + *at*

= 0 + (0.5)4 = 2 m/s

$s=ut+\frac{1}{2}a{t}^{2}$

$0.5=0+\frac{1}{2}a{\left(0.5\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow a=4\mathrm{m}/{\mathrm{s}}^{2}$

For the next half metre, *u* = 2 m/s, *a* = 4 m/s^{2} and *s* = 0.5.

$\Rightarrow 0.5=2t+\left(\frac{1}{2}\right)4{t}^{2}$

⇒ 2*t*^{2} + 2*t* − 0.5 = 0

⇒ 4*t*^{2} + 4*t* − 1 = 0

$\Rightarrow t=\frac{-4\pm \sqrt{16+16}}{2\times 4}\phantom{\rule{0ex}{0ex}}=\frac{1.656}{8}=0.2027$

Therefore, the time taken to cover the next half metre is 0.21 s.

#### Page No 97:

#### Answer:

Let

*f *be the applied force,

*R* be the normal reaction force and

*F* be the frictional force.

The coefficient of static friction is given by

$u=\mathrm{tan}\mathrm{\lambda}=\frac{\mathit{F}}{\mathit{R}}$

(where λ is the angle of friction)

When *F* = μ*R*, *F* is the limiting friction (maximum friction). When applied force increases and the body still remains still static then the force of friction increases up to its maximum value equal to limiting friction (μ*R*).

$F\mathit{<}\mu R\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}\mathrm{\lambda}=\frac{\mathit{F}}{\mathit{R}}\mathit{\le}\frac{\mathit{\mu}\mathit{R}}{\mathit{R}}$

⇒ tan λ ≤ μ

⇒ λ ≤ tan^{−1} μ

#### Page No 97:

#### Answer:

From the above diagrams:

*T *+ *ma* − *mg* = 0

*T* + 0.5*a* − 0.5 *g* = 0 (1)

μ*R* + *ma* + *T*_{1} −* T* = 0

μ*R* + 1*a* + *T*_{1} − *T* = 0 (2)

μ*R* + 1*a* − *T*_{1} = 0

μ*R* + *a* = *T*_{1} (3)

From Equations (2) and (3) we have

μ*R* + *a* =* T* − *T*_{1}

⇒ *T* − *T*_{1} = *T*_{1}

⇒ *T* = 2*T*_{1}

So, Equation (2) becomes

μ*R* + *a* + *T*_{1}_{ }− 2*T*_{1} = 0

⇒ μ*R* + *a* − *T*_{1} = 0

⇒ *T*_{1}_{ }= μ*R* + *a*

= 0.2*g* + *a* (4)

and Equation (1) becomes

2*T*_{1} + 0.5*a* − 0.5*g* = 0

$\Rightarrow {T}_{1}=\frac{0.5g-0.5a}{2}\phantom{\rule{0ex}{0ex}}=0.25g-0.25a\left(5\right)$

From Equations (4) and (5)

0.2*g* + *a* = 0.25*g* − 0.25*a*

$\Rightarrow a=\frac{0.05}{1.25}\times 10\phantom{\rule{0ex}{0ex}}=0.4\times 10m/{s}^{2}\left[g=10m/{s}^{2}\right]$

Therefore,

(a) the acceleration of each 1 kg block is 0.4 m/s^{2},

(b) the tension in the string connecting the 1 kg blocks is

*T*_{1} = 0.2*g* + *a *+ 0.4 = 2.4 N

and

(c) the tension in the string attached to the 0.5 kg block is

*T* = 0.5*g* − 0.5*a*

= 0.5 × 10 − 0.5 × 0.4

= 4.8 N.

#### Page No 98:

#### Answer:

From the free body diagram:

μ_{1}*R* + *m*_{1}*a* − *F* = 0

μ_{1}*R* + 1 − 16 = 0 (*R *= *mg *cos θ)

⇒ μ_{1}(2*g*) + (−15) = 0

${\mathrm{\mu}}_{1}=\frac{15}{20}=0.75$

Again,

μ_{2}*R*_{1} + *ma* = *F* − *mg* sin θ = 0

μ_{2}*R*_{1} + 4 × 0.5 = 16 − 4*g* sin 30° = 0

*R*_{1}_{ }= *mg *cos θ (θ = 30°)

$\Rightarrow {\mathrm{\mu}}_{2}\left(20\sqrt{3}\right)+2+16-20=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\mu}}_{2}=\frac{2}{20}\sqrt{3}=\frac{1}{17.32}\phantom{\rule{0ex}{0ex}}=0.057=0.06$

Therefore, the friction coefficients at the two contacts with blocks are

μ_{1} = 0.75 and μ_{2} = 0.06.

#### Page No 98:

#### Answer:

Consider that a 15 kg object is moving downward with an acceleration *a*.

From the above diagram,

*T* + *m*_{1}*a** *− *m*_{1}*g** = *0

*T* + 15*a* − 15*g* = 0

⇒ *T* = 15*g* − 15*a* (1)

Now,

*T*_{1} − *m*_{2}*g* − *m*_{2}*a* = 0

*T*_{1} − 5*g* − 5*a* = 0

⇒* **T*_{1}_{ }= 5*g* + 5*a* (2)

Again,

*T* − (*T*_{1} + 5*a* + *m*_{2}R) = 0

⇒ *T* − (5*g* + 5*a* + 5*a* +*m*_{2}R) = 0 (3)

(where *R* = μ*g*)

From Equations (1) and (2),

15*g* − 15*a* = 5*g* + 10*a* + 0.2 (5*g*)

⇒ 25*a* = 90 [*g* = 10 m/s^{2}]

⇒ *a* = 3.6 m/s^{2}

From Equation (3),

*T* = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 = 96 N in the left string.

From Equation (2),

*T*_{1} = 5*g* + 5*a*

= 5 × 10 + 5 × 36

= 50 + 18

= 68 N in the right string.

#### Page No 98:

#### Answer:

Given,

initial velocity of the vehicle, *u* = 36 km/h = 10 m/s

final velocity of the vehicle, *v* = 0

*s* = 5 m, $\mathrm{\mu}=\frac{4}{3}$, *g* = 10 m/s^{2}

Let the maximum angle of incline be $\theta $.

Using the equation of motion

$a=\frac{{v}^{2}-{u}^{2}}{2s}=\frac{0-{10}^{2}}{2\times 5}\phantom{\rule{0ex}{0ex}}=-10\mathrm{m}/{\mathrm{s}}^{2}$

From the free body diagram

*R* − *mg* cos θ = 0

⇒ *R* = *mg* cos θ (1)

Again,

*ma* + *mg* sin θ − μ *R* = 0

⇒ *ma* + *mg* sin θ − μ*mg* cos θ = 0

⇒ *a* + *g* sin θ − μ*g* cos θ = 0

$\Rightarrow 10+10\mathrm{sin}\mathrm{\theta}-\left(\frac{4}{3}\right)\times 10\mathrm{cos}\mathrm{\theta}=0$

⇒ 30 + 30 sin θ − 40 cos θ = 0

⇒ 3 + 3 sin θ − 4 cos θ = 0

⇒ 4 cos θ − 3 sin θ = 3

$\Rightarrow 4\sqrt{\left(1-{\mathrm{sin}}^{2}\mathrm{\theta}\right)}=3+3\mathrm{sin}\mathrm{\theta}$

On squaring, we get

16 (1 − sin^{2} θ) = 9 + 9 sin^{2} θ + 18 sin θ

25 sin^{2} θ + 18 sin θ − 7 = 0

$\Rightarrow \mathrm{sin}\mathrm{\theta}=\frac{18+\sqrt{{18}^{2}-4\left(25\right)\left(-7\right)}}{2\times 25}\phantom{\rule{0ex}{0ex}}=\frac{-18+32}{50}=\frac{14}{50}=0.28\left(\mathrm{Taking}\mathrm{positve}\mathrm{sign}\mathrm{only}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\theta}={\mathrm{sin}}^{-1}\left(0.28\right)=16\xb0$

Therefore, the maximum incline of the road, θ = 16°.

#### Page No 98:

#### Answer:

To reach the 50 m distance in minimum time, the superman has to move with maximum possible acceleration.

Suppose the maximum acceleration required is '*a*'.

∴ *ma* − μ*R* = 0 ⇒ *ma* = μ *mg*

⇒ *a* = μ*g* = 0.9 × 10 = 9 m/s^{2}

(a) As per the question, the initial velocity,

*u* = 0, *t* = ?

*a* = 9 m/s^{2}, *s* = 50 m

From the equation of motion,

$s=ut+\left(\frac{1}{2}\right)a{t}^{2}$

$50=0+\left(\frac{1}{2}\right)9{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{10}{3}\mathrm{s}$

(b) After covering 50 m, the velocity of the athelete is

*v* = *u *+ *at*

$=0+9\times \left(\frac{10}{3}\right)\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=30\mathrm{m}/\mathrm{s}$

The superman has to stop in minimum time. So, the deceleration, *a* = − 9 m/s^{2} (max)

*R* = *mg*

*ma* = μ*R* (maximum frictional force)

*ma* = μ*mg*

⇒ *a* = μ*g*

= 9 m/s^{2} (deceleration)

*u*_{1} = 30 m/s,* v* = 0

$\Rightarrow t=\frac{{v}_{1}-{u}_{1}}{a}\phantom{\rule{0ex}{0ex}}=\frac{0-30}{-a}\phantom{\rule{0ex}{0ex}}=\frac{-30}{-a}=\frac{10}{3}\mathrm{s}$

#### Page No 98:

#### Answer:

When the driver applies hard brakes, it signifies that maximum force of friction is developed between the tyres of the car and the road.

So, maximum frictional force = μ*R*

From the free body diagram,

*R* − *mg* cos θ = 0

⇒ *R* = *mg *cos θ (1)

and

μ*R* + *ma* − *mg* sin θ = 0 (2)

⇒ μ *mg* cos θ + *ma* − *mg *sin θ = 0

where θ = 30˚

$\Rightarrow \mathrm{\mu}g\mathrm{cos}\mathrm{\theta}+a-10\times \left(\frac{1}{2}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=5-\left[\frac{1}{2}\sqrt{3}\right]\times 10\left(\frac{\sqrt{3}}{2}\right)\phantom{\rule{0ex}{0ex}}=5-\frac{10\times 3}{4}\phantom{\rule{0ex}{0ex}}=\frac{20}{4}-\frac{10\times 3}{4}\phantom{\rule{0ex}{0ex}}=\frac{-10}{4}\phantom{\rule{0ex}{0ex}}=-2.5\mathrm{m}/{\mathrm{s}}^{2}$

*s* = 12.8 m

*u* = 6 m/s

∴ Velocity at the end of incline

$\nu =\sqrt{{u}^{2}+2as}\phantom{\rule{0ex}{0ex}}=\sqrt{{6}^{2}+2\left(2.5\right)\left(12.8\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{36+64}\phantom{\rule{0ex}{0ex}}=10\mathrm{m}/\mathrm{s}=36\mathrm{km}/\mathrm{h}$

Therefore, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h.

#### Page No 98:

#### Answer:

Let a be the maximum acceleration of the car for crossing the bridge.

From the above diagram,

*ma* = μ*R*

(For more accelerations the tyres will slip)

*ma* = μ*mg*

*a* = μ*g* = 1 × 10 = 10 m/s^{2}

To cross the bridge in minimum possible time, the car must be at its maximum acceleration.

*u* = 0, *s* = 500 m, *a *= 10 m/s^{2}

From the equation of motion,

$s=ut+\frac{1}{2}a{t}^{2}$

Substituting respective values

$500=0+\left(\frac{1}{2}\right)10{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{100}=10\mathrm{s}$

Therefore, if the car's acceleration is less than 10 m/s^{2}, it will take more than 10 s to cross the bridge. So, one cannot drive through the bridge in less than 10 s.

#### Page No 98:

#### Answer:

(a) From the free body diagram

*R* = 4*g* cos 30°

$\Rightarrow \mathrm{R}=4\times 10\times \frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}=20\sqrt{3}\mathrm{N}\left(1\right)$

μ_{2}*R* + *m*_{1}*a* − *p* − *m*_{1}*g* sin θ = 0

μ_{2}*R* + 4*a* − *p* − 4*g* sin 30° = 0

⇒ 0.3 $\times $ (40) cos 30° + 4*a* − *p* − 40 sin 30° = 20 (2)

*R*_{1} = 2*g* cos 30° $=10\sqrt{3}$ (3)

*p* + 2*a* − μ_{1}*R*_{1} − 2*g* sin 30° = 0 (4)

From Equation (2),

$6\sqrt{3}+4a-p-20=0$

From Equation (4),

$p+2a+2\sqrt{3}-10=10\phantom{\rule{0ex}{0ex}}6\sqrt{3}+6a+30+2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 6a=30-8\sqrt{3}\phantom{\rule{0ex}{0ex}}=30-13.85=16.15\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{16.15}{6}\phantom{\rule{0ex}{0ex}}=2.69=2.7\mathrm{m}/{\mathrm{s}}^{2}$

(b) In this case, the 4 kg block will move at a higher acceleration because the coefficient of friction is less than that of the 2 kg block. Therefore, the two blocks will move separately. By drawing the free body diagram of 2 kg mass, it can be shown that *a* = 2.4 m/s^{2}.

#### Page No 98:

#### Answer:

From the free body diagram

*R*_{1} = *M*_{1}*g* cos θ (1)

*R*_{2} = *M*_{2}*g* cos θ (2)

*T* + *M*_{1}*g* sin θ − *M*_{1}*a* − μ*R*_{1} = 0 (3)

*T* −* **M*_{2}*g* + *M*_{2}*a* + μ*R*_{2} = 0 (4)

From Equation (3),

*T* + *M*_{1}*g* sin θ − *M*_{1}_{ }*a* − μ*M*_{1}*g* cos θ = 0 (5)

From Equation (4),

*T* − *M*_{2}_{ }*g* sin θ + *M*_{2}_{ }*a* + μ*M*_{2}* g* cos θ = 0 (6)

From Equations (5) and (6),

*g *sin θ(*M*_{1} +* **M*_{2}) − *a*(*M*_{1} + *M*_{2}) − μ*g* cos θ(*M*_{1} + *M*_{2})

⇒ *a*(*M*_{1} + *M*_{2}) = *g* sin θ(*M*_{1} + *M*_{2}) = μ*g* cos θ(*M*_{1} + *M*_{2})

⇒ *a* = *g*(sin θ − μ cos θ)*a* − *g*(sin θ − μ cos θ)

∴ The acceleration of the block (system) = *g*(sin θ − μcos θ)

The force exerted by the rod on one of the blocks is tension, *T*.

*T* = −*M*_{1}*g** *sin θ + *M*_{1}*a* + μ*M*_{1}*g* cos θ

*T* = −*M*_{1}*g* sin θ +* **M*_{1}(*g* sin θ − μ*g* cos θ) + μ*M*_{1}*g* cos θ = 0

#### Page No 98:

#### Answer:

Let *P* be the force applied to slide the block at an angle θ.

From the free body diagram,

*R* + *P* sin θ − *mg* = 0

⇒ *R* = −*P* sin θ + *mg* (1)

μ*R* = *P* cos θ (2)

From Equation (1),

μ(*mg* − *P* sin θ)−*P** *cos θ = 0

⇒ μ*mg* = μ*P* sin θ + *P* cos θ

$\Rightarrow P=\frac{\mathrm{\mu}mg}{\mathrm{\mu}\mathrm{sin}\mathrm{\theta}+\mathrm{cos}\mathrm{\theta}}$

The applied force *P* should be minimum, when μ sin θ + cos θ is maximum.

Again, μ sin θ + cos θ is maximum when its derivative is zero:

$\frac{d}{d\mathrm{\theta}}\left(\mathrm{\mu}\mathrm{sin}\mathrm{\theta}+\mathrm{cos}\mathrm{\theta}\right)=0$

⇒ μ cos θ − sin θ = 0

θ = tan^{−1} μ

So, $P=\frac{\mathrm{\mu}mg}{\mathrm{\mu}\mathrm{sin}\mathrm{\theta}+\mathrm{cos}\mathrm{\theta}}$

Dividing numerator and denominator by cos θ, we get

$=\frac{\mathrm{\mu}mg/\mathrm{cos}\mathrm{\theta}}{{\displaystyle \frac{\mathrm{\mu}\mathrm{sin}\mathrm{\theta}}{\mathrm{cos}\mathrm{\theta}}}+{\displaystyle \frac{\mathrm{cos}\mathrm{\theta}}{\mathrm{cos}\mathrm{\theta}}}}$

$P=\frac{\mathrm{\mu}mg\mathrm{sec}\mathrm{\theta}}{1+\mathrm{\mu}\mathrm{tan}\mathrm{\theta}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\mu}mg\mathrm{sec}\mathrm{\theta}}{1+{\mathrm{tan}}^{2}\mathrm{\theta}}=\frac{\mathrm{\mu}mg}{\mathrm{sec}\mathrm{\theta}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\mu}mg}{\sqrt{\left(1+{\mathrm{tan}}^{2}\mathrm{\theta}\right)}}=\frac{\mathrm{\mu}mg}{\left(1+{\mathrm{\mu}}^{2}\right)}$(using the property $1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta $)

Therefore, the minimum force required is $\frac{\mathrm{\mu}mg}{\sqrt{\left(1+{\mathrm{\mu}}^{2}\right)}}$ at an angle θ = tan^{−1} μ.

#### Page No 98:

#### Answer:

Let *T* be the maximum force exerted by the man on the rope.

From the free body diagram,

*R* + *T* = *Mg*

⇒ *R* = *Mg* − *T* (1)

Again,

*R*_{1} − *R* − *mg* = 0

⇒ *R*_{1} = *R* + *mg* (2)

and

*T* − μ*R*_{1} = 0

From Equation (2),

*T *− μ(*R* + *mg*) = 0

⇒ *T* − μR − μ *mg *= 0

⇒ *T* − μ(*Mg* − T) − μ*mg* = 0

*T* − μ*Mg* + μ*t* − μ*mg* = 0

⇒ *T* (1 + μ) = μ*Mg* + μ*mg*

$\Rightarrow T=\frac{\mathrm{\mu}\left(M+m\right)g}{1+\mathrm{\mu}}$

Therefore, the maximum force exerted by the man is $\frac{\mathrm{\mu}\left(M+m\right)g}{1+\mathrm{\mu}}$.

#### Page No 98:

#### Answer:

Consider the free body diagram.

(a) For the mass of 2 kg, we have:

*R*_{1} − 2*g* = 0

⇒ *R*_{1} = 2 × 10 = 20

2*a* + 0.2 *R*_{1} − 12 = 0

⇒ 2*a* + 0.2 (20) = 12

⇒ 2*a* = 12 − 4

⇒ *a* = 4 m/s^{2}

Now,

4*a* − μ*R*_{1} = 0

⇒ 4*a* = μ*R*_{1} = 0.2 (20) = 4

⇒ *a*_{1} = 1 m/s^{2}

The 2 kg block has acceleration 4 m/s^{2} and the 4 kg block has acceleration 1 m/s^{2}.

(ii) We have:

*R*_{1} = 2*g* = 20

*Ma* = μ*R*_{1} = 0

*a* = 0

And,

*Ma* + μ*m*g − *F *= 0

4*a* + 0.2 × 2 × 10 − 12 = 0

⇒ 4*a* + 4 = 12

⇒ 4*a* = 8

⇒ *a *= 2 m/s^{2}

#### Page No 98:

#### Answer:

Given:

μ_{1} = 0.2

μ_{2} = 0.3

μ_{3} = 0.4

Using the free body diagram, we have:

(a) When the 10 N force is applied to the 2 kg block, it experiences maximum frictional force.

Here,

μ_{1}*R*_{1} = μ_{1} × *m*_{1}*g*

μ_{1}*R*_{1} = μ_{1} × 2*g* = (0.2) × 20

= 4 N (From the 3 kg block)

Net force experienced by the 2 kg block = 10 − 4 = 6 N

∴ ${a}_{1}=\frac{6}{2}=3\mathrm{m}/{\mathrm{s}}^{2}$

But for the 3 kg block (Fig. 3), the frictional force from the 2 kg block, i.e, 4 N, becomes the driving force and the maximum frictional force between the 3 kg and 7 kg blocks.

Thus, we have:

μ_{2} = *R*_{2} = μ_{2}*m*_{2}*g* = (0.3) × 5 kg

= 15 N

Therefore, the 3 kg block cannot move relative to the 7 kg block.

The 3 kg block and the 7 kg block have the same acceleration (*a*_{2} = *a*_{3}), which is due to the 4 N force because there is no friction from the floor.

$\therefore {a}_{2}={a}_{3}=\frac{4}{10}=0.4\mathrm{m}/{\mathrm{s}}^{2}$

(b) When the 10 N force is applied to the 3 kg block, it experiences maximum frictional force of (15 + 4) N, i.e., 19 N, from the 2 kg block and the 7 kg block.

So, it cannot move with respect to them.

As the floor is frictionless, all the three bodies move together.

$\therefore {a}_{1}={a}_{2}={a}_{3}=\frac{10}{12}=\left(\frac{5}{6}\right)\mathrm{m}/{\mathrm{s}}^{2}$

(c) Similarly, it can be proved that when the 10 N force is applied to the 7 kg block, all three blocks move together with the same acceleration.

$\therefore {a}_{1}={a}_{2}={a}_{3}=\left(\frac{5}{6}\right)\mathrm{m}/{\mathrm{s}}^{2}$

#### Page No 98:

#### Answer:

From the free body diagrams of the two blocks, we have

*R*_{1} = *mg ...*(i)

*F* = *μR*_{1}+*T ...*(ii)

*T* − *μR*_{1} = 0 ..(iii)

From equations (i) and (ii), we have

*F* − *μmg* = *T* ...(ii)

From equations (i) and (iii), we have

*T* = *μmg*

Putting *T* = *μmg* in equation (ii), we have

*F* = *μmg* + *μmg* = 2*μmg*

(b) From the free body diagram of upper block, we have

2*F* − *T* − *μmg* = *ma* ....(i)

From the free body diagram of lower block, we have

*T* *= Ma* + *μmg*

Putting the value of *T* in (i), we get

2*F* − *Ma* − *μmg* − *μmg* = *ma*

Putting *F* = 2*μmg**, *we get

2(2*μ**mg*) − 2*μmg* = *a*(*M* + *m*)

⇒ 4*μmg* − 2*μmg* = *a*(*M* + *m*)

$\Rightarrow a=\frac{2\mathrm{\mu}mg}{M+m}$ in opposite directions.

#### Page No 99:

#### Answer:

From the free body diagram, we have:

*R*_{1} + *ma* − *mg* = 0

⇒ *R*_{1} = *m* (*g* − *a*)

= *mg* − *ma* ...(i)

Now,

*F* − *T* − μ*R*_{1} = 0 and

*T* − μ*R*_{1} = 0

⇒ *F* − [μ (*mg* − *ma*)] − μ(*mg* − *ma*) = 0

⇒ *F* − μ* mg* − μ*ma* − μ*mg* + μ*ma* = 0

⇒ *F* = 2 μ*mg* − 2 μ*ma*

= 2 μ*m* (*g* − *a*)

(b) Let the acceleration of the blocks be *a*_{1}.

*R*_{1} = *mg* − *ma* ....(i)

And,

2*F** *−*T* − μ*R*_{1} = *ma*_{1} ...(ii)

Now,

*T* = μ*R*_{1} + *Ma*_{1}

= μ*mg* − μ*ma* + *Ma*_{1}

Substituting the value of *F* and *T* in equation (ii), we get:

2[2μ*m*(*g* − *a*)] − (μ*mg* − μ*ma* + *Ma*_{1}) − μ*mg* + μ*ma* = *ma*_{1}

⇒ 4μ*mg** *− 4μ*ma* − 2μ*mg* + 2μ*ma*= *ma*_{1} + *Ma*_{1}

$\Rightarrow {a}_{1}=\frac{2\mathrm{\mu}m\left(g-a\right)}{M+m}$

Thus, both the blocks move with same acceleration *a*_{1} but in opposite directions.

#### Page No 99:

#### Answer:

From the free body diagram:

*R*_{1} + *QE* − *mg* = 0

where

* **R*_{1}_{ }is the normal reaction force

*Q* is the charge

*E* is the small electric field

⇒ *R*_{1} = *mg* − *QE* (1)

*F* − *T *− μ*R*_{1}* = *0

*F* − *T *= μ*R*_{1}

where *F *is the maximum horizontal force required

From Equation (1),

*F − T *− μ(*mg* − *QE*) = 0

*F − T *= μ(*mg* − *QE*)

⇒ *F − T* − μ*mg* + μ*QE* = 0 (2)

*T* − μR_{1} = 0

⇒ *T* = μR_{1} = μ (*mg* − *QE*) (3)

From Equation (2),

*F* − μ*mg* + μ *QE* − μ*mg* + μ*QE* = 0

⇒ *F* − 2 μ*mg* + 2μ*QE* = 0

⇒* F* = 2μ*mg* − 2μ*QE*

⇒ *F *= 2μ (*mg* − *QE*)

Therefore, the maximum horizontal force that can be applied is 2μ (*mg* −* QE*).

#### Page No 99:

#### Answer:

When a block slips on a rough horizontal table, the maximum frictional force acting on it can be found from the free body diagram (see below).

*R* = *mg*

(i) (ii)

From the free body diagram,

*F* − μ*R* = 0

⇒ *F* = μ*R* = μ*mg*

But the table is at rest, so the frictional force at the legs of the table is also μ*R*. Let this be *f*, so from the free body diagram

*f* − μ*R** *= 0

⇒ *f* = μ*R* = μ*mg*

Therefore, the total frictional force on the table by the floor is μ*mg*.

#### Page No 99:

#### Answer:

Let us the acceleration of the block of mass *M* be *a* and let it be towards right. Therefore, the block of mass *m* must go down with acceleration 2*a*. As both the blocks are* *in contact, it (block of mass m) will also have acceleration *a* towards right. Hence, it will experience two inertial forces as shown in the free body diagram given below.

(Free body diagram 1)

From the free body diagram 1, we have:

(Free body diagram-2)

*R*_{1} − *ma* = 0

⇒ *R*_{1} = *ma* ....(i)

Again,

2*ma* + *T* − *Mg* + μ_{1}*R*_{1} = 0

⇒ *T* = *Mg* − (2 + μ_{1}) *ma* ....(ii)

Using the free body diagram 2, we have:

*T* + μ_{1}*R*_{1} + *M*g − *R*_{2} = 0

Substituting the value of *R*_{1} from (i), we get:

*R*_{2} = *T* + μ_{1} *ma* + *mg*

Substituting the value of *T* from (ii), we get:

*R*_{2} = (*Mg* − 2*ma* − μ_{1}*ma*) = μ_{1} *ma* + *Mg* + *ma*

∴ *R*_{2} = *Mg* + *Ma* − 2*ma* ....(iii)

Again using the free body diagrams −2,

*T* + *T* − *R* − *Ma* − μ_{2}*R*_{2} = 0

⇒ 2*T** *− *Ma* − *ma* − μ2(*Mg* + *mg* − 2*ma*) = 0

Substituting the values of *R*_{1} and *R*_{2} from (i) and (iii), we get:

2*T* = (*M* + *m*)*a* + μ_{2} (*Mg* + *mg* − 2*ma*) ....(iv)

From equations (ii) and (iv), we have:

2*T* = 2*mg* − 2(2 + μ_{1}) *ma*

= (*M* + *m*) *a* + μ_{1}(*Mg* + *mg* − 2*ma*)

⇒ 2*mg* − μ_{2} (*M* + *m*)*g* = *a*[*M* + *m* − 2μ_{2}*m* + 4*m* + 2μ_{1}*m*]

Therefore, the acceleration of the block of mass *M* in the given situation is given by

$a=\frac{\left[2m+{m}_{2}\left(M+m\right)\right]g}{M+m\left[5+2\left({\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}\right)\right]}$

#### Page No 99:

#### Answer:

Net force on the block$=\sqrt{\left({20}^{2}+{15}^{2}\right)}-\left(0.5\right)\times 40$

= 25 − 20 = 5 N

$\therefore \mathrm{tan}\mathrm{\theta}=\frac{20}{15}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta ={\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)=53\xb0$

Therefore, the block will move at 53° angle with the 15 N force.

#### Page No 99:

#### Answer:

Thus, we have:

μ*R* + μ*R* = *m*g

⇒ 2μ*R* = 40 × 10

$\Rightarrow R=\frac{40\times 10}{2\times 0.8}=250\mathrm{N}$

Normal force = 250 N

#### Page No 99:

#### Answer:

Let *a*_{1} and *a*_{2} be the accelerations of masses *m* and *M*, respectively.

Also, *a*_{1} > *a*_{2} so that mass *m* moves on mass *M*.

Let after time *t*, mass *m* is separated from mass *M*.

Using the equation of motion

During this time, mass *m* covers $vt+\frac{1}{2}{a}_{1}{t}^{2}$ and ${\mathrm{s}}_{m}=vt+\frac{1}{2}{a}_{2}{t}^{2}$.

For mass *m* to separate from mass *M*, we have:

$vt+\frac{1}{2}{a}_{1}{t}^{2}=vt+\frac{1}{2}{a}_{2}{t}^{2}+1....\left(\mathrm{ii}\right)$

From the free body diagram, we have:

$m{a}_{1}+\frac{\mathrm{\mu}}{2}R=0\phantom{\rule{0ex}{0ex}}\Rightarrow m{a}_{1}=-\left(\frac{\mathrm{\mu}}{2}\right)mg=\left(\frac{\mathrm{\mu}}{2}\right)m\times 10\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{1}=-5\mathrm{\mu}$

Again,

$M{a}_{2}+\mathrm{\mu}\left(M+m\right)g-\left(\frac{\mathrm{\mu}}{2}\right)mg=0$

⇒ 2*Ma*_{2} + 2μ (*M* + *m*)*g* − μ*mg* = 0

⇒ 2*Ma*_{2} = μ*mg* − 2μ*mg* − 2μ*m**g*

$\Rightarrow {a}_{2}=\frac{-\mathrm{\mu}mg-2\mu Mg}{2M}$

Substituting the values of *a*_{1} and *a*_{2} in equation (i), we get:

$t=\sqrt{\frac{4Ml}{\left(M+m\right)\mathrm{\mu}g}}$

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