HC Verma Concepts Of Physics PDF Exercise Solutions Of Chapter 6 (Friction)

Solutions of H.C. Verma’s Concepts of Physics chapter 6 (Friction) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 6 (Friction)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 6 - Friction:

EXERCISE

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?

Sol:

Given: acceleration, a = – 0.4 m/s².

Let mass of the body be ‘m’ and F_f → frictional force.

From the free body diagram,

Vertical motion: R – mg = 0 → R = mg ……… (1)

Horizontal motion: ma = F_f = μR ……. (2)

From equation (1) and (2)

We get, μ = a/g = 4/10 = 0.4

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

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?

Sol:

Let mass of the body be ‘m’ and deceleration be a.

F_f → frictional force; μ = 0.10.

From the free body diagram,

Vertical motion: R – mg = 0 → R = mg ……… (1)

Horizontal motion: ma = F_f = μR ……. (2)

From equation (1) and (2)

We get, a = μg = 0.10 * 10 = 1 m/s²

Initial velocity u = 10 m/s; Final velocity v = 0 m/s; a = – 1 m/s² (deceleration).

We have, 2as = v² – u²

Or, s = (v² – u²)/2a = (0² – 10²)/2(– 1) = 50 m

So, it will travel 50 m before coming to rest. 

3. 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. 

Sol:

Body is kept on the horizontal table.

If no force is applied, no frictional force will be there

F_s → frictional force

F_a → Applied force

From graph it can be seen that when applied force is zero, frictional force is zero.

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.

Sol:

Given: distance covered, s = 8 m; time taken, t = 2 s; initial velocity, u = 0.

We know, s = ut + ½ at²

Or, 8 = 0 * 2 + ½ * a * 2² → a = 4 m/s².

Frictional force, F_f = μR

Let, the mass of body be ‘m’.

The equation of motion of the body along perpendicular to the plane: acceleration, a = 0.

→ R – mg cos 30⁰ = 0 → R = mg cos 30⁰ ------- (1)

The equation of motion of the body along the plane: acceleration, a = 4 m/s².

→ mg sin 30⁰ – F_f = ma

Or, mg sin 30⁰ – μR = 4m ------- (2)

Solving equation (1) and (2)

We get, mg sin 30⁰ – μmg cos 30⁰ = 4m

Or, μ = (g sin 30⁰ – 4)/g cos 30⁰ = 0.11

Therefore, the coefficient of kinetic friction between the two is 0.11.

5. 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.

Sol:

Given: mass of block, m = 5 kg; Force applied, F = 4 N; μ = 0.11.

The equation of motion of the body along perpendicular to the plane: acceleration, a = 0.

→ R – mg cos 30⁰ = 0 → R = 4g cos 30⁰ ------- (1)

The equation of motion of the body along the plane: acceleration, a.

→ mg sin 30⁰ – F_f + F = ma

Or, 4g * ½ – μR + 4 = 4a ------- (2)

Solving equation (1) and (2)

We get, 4g * ½ – 4* 0.11* g cos 30⁰ + 4 = 4a

Or, a = 5.04 ~ 5 m/s²  

Now, we have u = 0; t = 2 s; a = 5 m/s²

We know, s = ut + ½ at² = 0*2 + ½ * 5 * 2²

Or, s = 10 m

Therefore, the block move in the first two seconds after starting from rest is 10 m.

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