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

Board ISC and other board
Publications Bharti Bhawan Publishers
Ch-6 Friction
Class 11
Vol  1st
writer H C Verma
Book Name Concept of Physics
Topics Solution of Exercise Questions
Page-Number 97, 98, 99

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Question for Short Answer

Objective-I

Objective-II

Exercise


Friction Exercise Questions Solutions

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

Page-97

Question-1 :-

A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s2. What is the coefficient of kinetic friction between the block and the plane?

Answer-1 :-

Let m be the mass of the body.

A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s2. What is the coefficient of kinetic friction between the block and the plane? 1

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 − μkR = 0
(where μk is the coefficient of kinetic friction and a is deceleration)
or ma = μkR
From Equation (1),
ma =  μkmg
⇒ a = μkg
⇒ 4 = μkg

A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s2. What is the coefficient of kinetic friction between the block and the plane?2

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

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.

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,
μkR + ma − mg sin θ = 0
(where μis the coefficient of kinetic friction)
From Equation (1):
μkmg cos θ + ma − mg sin θ = 0
⇒ m (μkg cos θ + a − g sin θ) = 0
⇒ μk × 10 × cos 30° = g sin 30° − a

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. 2

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

Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest? The mass of the block is 4 kg.

From the above diagram:
F − ma − μkR + mg sin 30° = 0
4 − 4a − μkR + 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

Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest? The mass of the block is 4 kg. 2

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. img 2

= 6.41 N

friction Exercise img 12
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

friction Exercise img 13

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

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?

From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ                      (1)
and
ma + mg sin θ − μR = 0

friction Exercise img 18

 

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

friction Exercise img 19

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

friction Exercise img 20

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

friction Exercise img 22

(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).

friction Exercise img 23

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.

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. img 1

Answer-11 :-

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. img 2

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

friction Exercise img 26

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

friction Exercise img 27

= 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.

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. img 1

Answer-12 :-

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

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. img 2

friction Exercise img 29

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

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.

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