Friction HC Verma Exercise Questions Solutions Ch-6

Friction HC Verma Exercise Questions Solutions Ch-6 Concept of Physics Vol-1 for ISC Class-11. Step by Step Solution of Exercise Questions of Ch-6 Friction HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Friction Exercise Questions Solutions HC Verma Ch-6 Concept of Physics Vol-1 for ISC Class-11

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Friction Exercise Questions Solutions

(HC Verma Ch-6 Concept of Physics Vol-1 for ISC Class-11)

Page-97

Question-1 :-

Answer-1 :-

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_kR = 0
(where μ_k is the coefficient of kinetic friction and a is deceleration)
or ma = μ_kR
From Equation (1),
ma =  μ_kmg
⇒ a = μ_kg
⇒ 4 = μ_kg

⇒

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

Question-2 :-

A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest?

Answer-2 :-

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_kR = 0
(where μ_k is the coefficient of kinetic friction)
From Equation  …..   (1),
⇒ ma = μ_kmg
⇒ a = μ_kg = 0.1 × 10
= 1 m/s²
Given:
initial velocity, u = 10 m/s
final velocity, v = 0 m/s (block comes to rest)
a = −1 m/s² (deceleration)
Using equation of motion v² – u² = 2as
(where s is the distance travelled before coming to rest)

On substituting the respective values, we get

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

Question-3 :-              (Friction HC Verma Exercise)

A block of mass m is kept on a horizontal table. If the static friction coefficient is μ, find the frictional force acting on the block.

Answer-3 :-

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.

Question-4 :-

A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two.

Answer-4 :-

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

Again,
μ_kR + ma − mg sin θ = 0
(where μ_k is the coefficient of kinetic friction)
From Equation (1):
μ_kmg cos θ + ma − mg sin θ = 0
⇒ m (μ_kg cos θ + a − g sin θ) = 0
⇒ μ_k × 10 × cos 30° = g sin 30° − a

From the above diagram:
F − ma − μ_kR + mg sin 30° = 0
4 − 4a − μ_kR + 4g sin 30° = 0               (1)
R − 4g cos 30° = 0                               (2)
⇒ R = 4g cos 30° = 0
Substituting the values of R in Equation (1) we get
4 − 4a − 0.11 × 4g cos 30° + 4g sin 30° = 0

4 − 4a − 3.81 + 20 = 0
4 − 4a − 3.18 + 20 = 0
a ≈ 5 m/s²
For the block, u = 0, t = 2 s and a = 5 m/s².
According to the equation of motion,

Therefore, the block will move 10 m.

Question-6 :-

A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.

Answer-6 :-

(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 = μR + 2g sin 30°          (1)
(where μ is the coefficient of static friction)
R = mg cos 30°
Substituting the respective values in Equation (1), we get

3.39 + 9.8 ≈ 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 = 2g sin 30° − μR

= 6.41 N

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

g = 10 m/s², 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 + μR − F cos θ = 0
⇒ mg sin θ + μ(mg cos θ + F sin θ) − F cos θ = 0
⇒ mg sin θ + μmg cos θ + μF sin θ − F cos θ = 0

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

Question-8 :-            (Friction HC Verma Exercise)

In a children-park an inclined plane is constructed with an angle of incline 45° in the middle part (in the following figure). Find the acceleration of boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 19 m/s².

Answer-8 :-

Let m be the mass of the boy.

From the above diagram:
R − mg cos 45° = 0
R = mg cos 45° =mg/√2     ………….        (1)
Net force acting on the boy, making him slide down
= mg sin 45° − μR
= mg sin 45° − μmg cos 45°

Question-9 :-

A body starts slipping down an incline and moves half metre in half second. How long will it take to move the next half metre?

Answer-9 :-

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 − μR = 0

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

⇒ a = 4 m/s²
For the next half metre, u = 2 m/s, a = 4 m/s² and s = 0.5.

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

Question-10 :-

The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that, if λ be the angle of friction and μ the coefficient of static friction λ ≤ tan⁻¹ μ.

Answer-10 :-

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

(where λ is the angle of friction)
When F = μR, 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 (μR).

Question-11 :-            (Friction HC Verma Exercise)

Consider the situation shown in the following figure. Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.

Answer-11 :-

From the above diagrams:
T + ma − mg = 0
T + 0.5a − 0.5 g = 0                       (1)
μR + ma + T₁ − T = 0
μR + 1a + T₁ − T = 0                      (2)
μR + 1a − T₁ = 0
μR + a = T₁                                    (3)
From Equations (2) and (3) we have
μR + a = T − T₁
⇒ T − T₁ = T₁
⇒ T = 2T₁
So, Equation (2) becomes
μR + a + T₁ − 2T₁ = 0
⇒ μR + a − T₁ = 0
⇒ T₁ = μR + a
= 0.2g + a                       (4)

and Equation (1) becomes
2T₁ + 0.5a − 0.5g = 0

= 0.25g – 0.25a                    (5)
From Equations (4) and (5)
0.2g + a = 0.25g − 0.25a

= 0.4 x 10 m/s²                     [g = 10 m/s²] Therefore,
(a) the acceleration of each 1 kg block is 0.4 m/s²,
(b) the tension in the string connecting the 1 kg blocks is
T₁ = 0.2g + a + 0.4 = 2.4 N
​    and
(c) the tension in the string attached to the 0.5 kg block is
T = 0.5g − 0.5a
= 0.5 × 10 − 0.5 × 0.4
= 4.8 N.

Question-12 :-

If the tension in the string in the following figure is 16 N and the acceleration of each block is 0.5 m/s², find the friction coefficients at the two contact with the blocks.

Answer-12 :-

From the free body diagram:
μ₁R + m₁a − F = 0
μ₁R + 1 − 16 = 0 (R = mg cos θ)

Therefore, the friction coefficients at the two contacts with blocks are μ₁ = 0.75 and μ₂ = 0.06.

Question-13 :-

The friction co-efficient between the table and the block shown in the following figure is 0.2. Find the tensions in the two strings.

Answer-13 :-

Consider that a 15 kg object is moving downward with an acceleration a.
From the above diagram,
T + m₁a − m₁g = 0
T + 15a − 15g = 0
⇒ T = 15g − 15a                  (1)
Now,
T₁ − m₂g − m₂a = 0
T₁ − 5g − 5a = 0
⇒ T₁ = 5g + 5a                                   (2)
Again,
T − (T₁ + 5a + m₂R) = 0
⇒ T − (5g + 5a + 5a +m₂R) = 0        (3)
(where R = μg)
From Equations (1) and (2),
15g − 15a = 5g + 10a + 0.2 (5g)
⇒ 25a = 90          [g = 10 m/s²] ⇒ a = 3.6 m/s²
From Equation (3),
T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 = 96 N in the left string.
From Equation (2),
T₁ = 5g + 5a
= 5 × 10 + 5 × 36
= 50 + 18
= 68 N in the right string.

Question-14 :-

The friction coefficient between a road and the type of a vehicle is 4/3. Find the maximum incline the road may have so that once had brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.

Answer-14 :-

Given,
initial velocity of the vehicle, u = 36 km/h = 10 m/s
final velocity of the vehicle, v = 0
s = 5m, μ = 4/3, g = 10 m/s²
Let the maximum angle of incline be θ.
Using the equation of motion

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 − μmg cos θ = 0
⇒ a + g sin θ − μg cos θ = 0

⇒ 30 + 30 sin θ − 40 cos θ = 0
⇒ 3 + 3 sin θ − 4 cos θ = 0
⇒ 4 cos θ − 3 sin θ = 3

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

Question-15 :-        (Friction HC Verma Exercise)

The friction coefficient between an athelete’s shoes and the ground is 0.90. Suppose a superman wears these shoes and races for 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time that he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop?

Answer-15 :-

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 − μR = 0 ⇒ ma = μ mg
⇒ a = μg = 0.9 × 10 = 9 m/s²
(a) As per the question, the initial velocity,
u = 0, t = ?
a = 9 m/s², s = 50 m
From the equation of motion,

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

The superman has to stop in minimum time. So, the deceleration, a = − 9 m/s² (max)
R = mg
ma = μR                   (maximum frictional force)
ma = μmg
⇒ a = μg
= 9 m/s²  (deceleration)
u₁ = 30 m/s, v = 0

Question-16 :-

A care is going at a speed of 21.6 km/hr when it encounters at 12.8 m long slope of angle 30° (in the following figure). The friction coefficient between the road and the tyre is 1/2√3. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take g = 10 m/s².

Answer-16 :-

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 = μR
From the free body diagram,
R − mg cos θ = 0
⇒ R = mg cos θ                              (1)
and
μR + ma − mg sin θ = 0                   (2)
⇒ μ mg cos θ + ma − mg sin θ = 0

where θ = 30˚

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.

Question-17 :-

A car starts from rest on a half kilometre long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one cannot drive through the bridge in less than 10 s.

Answer-17 :-

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

From the above diagram,
ma = μR
(For more accelerations the tyres will slip)
ma = μmg
a = μg = 1 × 10 = 10 m/s²
To cross the bridge in minimum possible time, the car must be at its maximum acceleration.
u = 0, s = 500 m, a = 10 m/s²
From the equation of motion,

Therefore, if the car’s acceleration is less than 10 m/s², it will take more than 10 s to cross the bridge. So, one cannot drive through the bridge in less than 10 s.

Question-18 :-          (Friction HC Verma Exercise)

In the following figure shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is μ₁, and that between the block of mass 4.0 kg and incline is μ₂. Calculate the acceleration of the 2.0 kg block if (a) μ₁ = 0.20 and μ₂ = 0.30, (b) μ₁ = 0.30 and μ₂ = 0.20. Take g = 10 m/s².

Answer-18 :-

(a) From the free body diagram

R = 4g cos 30°

μ₂R + m₁a − p − m₁g sin θ = 0
μ₂R + 4a − p − 4g sin 30° = 0
⇒ 0.3 x (40) cos 30° + 4a − p − 40 sin 30° = 20             (2)

R₁ = 2g cos 30° = 10√3                       (3)
p + 2a − μ₁R₁ − 2g sin 30° = 0               (4)
From Equation (2),

= 2.69 = 2.7 m/s²

^{(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/s².}

Question-19 :-

Two masses M₁ and M₂ are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is μ. Find the acceleration of the system and the force by the rod on one of the blocks.

Answer-19 :-

^{From the free body diagram}

^{R₁ = M₁g cos θ                                             (1)
R₂ = M₂g cos θ                                             (2)
T + M₁g sin θ − M₁a − μR₁ = 0                      (3)
T − M₂g + M₂a + μR₂ = 0                             (4)
From Equation (3),
T + M₁g sin θ − M₁ a − μM₁g cos θ  = 0         (5)
From Equation (4),
T − M₂ g sin θ + M₂ a + μM₂ g cos θ = 0        (6)
From Equations (5) and (6),
g sin θ(M₁ + M₂) − a(M₁ + M₂) − μg cos θ (M₁ + M₂)
⇒ a(M₁ + M₂) = g sin θ(M₁ + M₂) = μg cos θ (M₁ + M₂)
⇒ 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 = −M₁g sin θ + M₁a + μM₁g cos θ
T = −M₁g sin θ + M₁(g sin θ − μg cos θ) + μM₁g cos θ = 0}

Question-20 :-

A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is μ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?

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