Question 1

A wooden block of mass m rests on a rough horizontal table (coefficient of friction = μ). It is pulled by a force F applied at an angle θ above the horizontal. The acceleration of the block (moving horizontally) is:

Accepted Answer

(F/m)(cos θ + μ sin θ) − μg. The upward component of F reduces the normal reaction: N = mg − F sinθ. Net horizontal force = F cosθ − μN, so a = (F/m)(cos θ + μ sin θ) − μg.

Question 2

Two blocks m₁ = 4 kg and m₂ = 2 kg, connected by a weightless rod, are placed on a plane inclined at 37°. The coefficients of dynamic (kinetic) friction of both m₁ and m₂ with the inclined plane are μ = 0.25. The common acceleration of the two blocks and the tension in the rod are:

Accepted Answer

4 m/s², T = 0. Since both blocks have the same μ, each would slide with the same acceleration a = g(sin37° − μcos37°) = 10(0.6 − 0.25×0.8) = 4 m/s², so the connecting rod carries no load and T = 0.

Question 3

A plank of mass m₁ = 8 kg with a bar of mass m₂ = 2 kg placed on its rough upper surface lies on the smooth floor of an elevator ascending with acceleration g/4. The coefficient of friction between m₁ and m₂ is μ = 1/5. A horizontal force F = 30 N is applied to the plank. The accelerations of the bar and the plank in the reference frame of the elevator are:

Accepted Answer

2.5 m/s², 25/8 m/s². In the elevator frame g_eff = g + g/4 = 12.5 m/s². Friction on the bar f = μ m₂ g_eff = (1/5)(2)(12.5) = 5 N, giving bar acceleration a₂ = 5/2 = 2.5 m/s²; the plank has a₁ = (F − f)/m₁ = (30 − 5)/8 = 25/8 m/s².

Question 4

The upper portion of an inclined plane of inclination α is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to the rough length is m : n, the coefficient of friction is:

Accepted Answer

((m + n)/n) tan α. The particle starts and ends at rest, so total gravity work equals friction work: mg sinα(L_smooth + L_rough) = μ mg cosα·L_rough. With L_smooth : L_rough = m : n, μ = ((m + n)/n) tan α.

Question 5

A fixed wedge has both its surfaces inclined at 45° to the horizontal. A particle P of mass m is held on the smooth face by a light string which passes over a smooth pulley at the top and is attached to a particle Q of mass 3m resting on the rough face. The system is released from rest. Given that the acceleration of each particle has magnitude g/(5√2), the tension in the string is:

Accepted Answer

6mg/(5√2). For the system: 3mg sin45° − μ·3mg cos45° − mg sin45° = 4m·a gives a = g/(5√2). For P alone, T − mg sin45° = m·a, so T = mg/√2 + mg/(5√2) = 6mg/(5√2).

Question 6

A plank of weight Mg is held at an angle α to the horizontal on two fixed supports A and B. The plank can slide against the supports without friction because of its weight. With what acceleration, and in what direction, should a man of mass m move along the plank so that the plank does not move?

Accepted Answer

g sin α (1 + M/m), down the incline. For the plank to stay at rest, the friction the man exerts must balance Mg sinα. Applying Newton's law to the man along the incline, mg sinα + Mg sinα = m·a, giving a = g sin α (1 + M/m) directed down the incline.

Question 7

STATEMENT-1: It is easier to pull a heavy object than to push it on level ground. STATEMENT-2: The magnitude of the frictional force depends on the nature of the two surfaces in contact.

Accepted Answer

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1. Both statements are true, but Statement-2 is not the correct explanation. Pulling is easier because it decreases the normal reaction (and hence friction), whereas pushing increases it — not merely because friction depends on the surfaces.

Question 8

A block of mass m₁ = 1 kg and another of mass m₂ = 2 kg are placed together on an inclined plane of angle of inclination θ. The coefficient of friction between m₁ and the plane is always zero; the coefficients of static and dynamic friction between m₂ and the plane are equal to μ = 0.3. Match the angle in List-I with the expression for the friction on block m₂ in List-II (g = acceleration due to gravity). List-I: P. θ = 5°, Q. θ = 10°, R. θ = 15°, S. θ = 20°. List-II: 1. m₂g sinθ, 2. (m₁ + m₂)g sinθ, 3. μ m₂g cosθ, 4. μ(m₁ + m₂)g cosθ.

Accepted Answer

P-2, Q-2, R-3, S-3. Slipping begins when tanθ = μ m₂/(m₁ + m₂) = 0.3×2/3 = 0.2, i.e. θ ≈ 11.5°. For θ = 5° and 10° friction is static and equals (m₁ + m₂)g sinθ (P-2, Q-2); for θ = 15° and 20° friction is kinetic and equals μ m₂g cosθ (R-3, S-3).

Question 9

A smooth block is released from rest on a 45° incline and slides a distance d. On a rough incline the time taken to slide the same distance is n times that on the smooth incline. The coefficient of friction is:

Accepted Answer

μₖ = 1 − 1/n². Smooth: a₁ = g sinθ; rough: a₂ = g(sinθ − μₖcosθ). Equal distance with t₂ = n·t₁ requires a₁ = n²a₂, so at 45° (1 − μₖ) = 1/n², giving μₖ = 1 − 1/n².

Question 10

The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough. A body starting from rest at the top comes to rest again at the bottom if the coefficient of friction for the lower half is:

Accepted Answer

2 tan θ. By the work-energy theorem, the potential energy lost over the whole length L equals the friction work over the lower (rough) half only: mg sinθ·L = μ mg cosθ·(L/2), giving μ = 2 tan θ.

Question 11

The minimum force required to start pushing a body up a rough (friction coefficient μ) inclined plane is F₁, while the minimum force needed to prevent it from sliding down is F₂. If the inclined plane makes an angle θ with the horizontal such that tan θ = 2μ, then the ratio F₁/F₂ is:

Accepted Answer

3. F₁ = mg sinθ + μmg cosθ and F₂ = mg sinθ − μmg cosθ, so F₁/F₂ = (tanθ + μ)/(tanθ − μ) = (2μ + μ)/(2μ − μ) = 3.

Question 12

The frictional force is -

Accepted Answer

Self adjustable. Static friction is a self-adjustable force: it adjusts itself to balance the applied force up to the limiting value.

Question 13

A block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them. Consider the statements: (a) The graph is a straight line of slope 45°. (b) The graph is a straight line parallel to the F axis. (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. Which of the following is correct?

Accepted Answer

c, d. Up to the limiting force friction balances F, so the graph is a 45° straight line; a small kink appears as motion begins, after which f stays constant and the graph runs parallel to the F-axis. Hence statements (c) and (d) are correct.

Question 14

A block A kept on an inclined surface just begins to slide if the inclination is 30°. The block is replaced by another block B and it is found that it just begins to slide if the inclination is 40° -

Accepted Answer

all the three are possible. The angle of repose gives μ = tan θ and is independent of the block's mass, so the sliding angles reveal nothing about the masses — all three cases are possible.

Question 15

It is easier to pull a body than to push, because-

Accepted Answer

the friction force is more in pushing than that in pulling. Pushing has a downward force component that raises the normal reaction and hence the friction, so the friction force is greater when pushing than when pulling.

Question 16

The coefficient of static friction between two surfaces depends on -

Accepted Answer

Nature of surfaces. The coefficient of static friction depends only on the nature of the surfaces in contact; it is independent of the contact area.

Question 17

A box is lying on an inclined plane. If the box starts sliding when the angle of inclination is 60°, then the coefficient of static friction of the box and plane is-

Accepted Answer

1.732. At the angle of sliding, μ = tan θ = tan 60° = √3 ≈ 1.732.

Question 18

A 20 kg block is initially at rest. A 75 N force is required to set the block in motion. After the motion, a force of 60 N is applied to keep the block moving with constant speed. The coefficient of static friction is-

Accepted Answer

0.35. The 75 N needed to start motion equals the limiting static friction, so μs = 75/(20 × g) ≈ 0.35.

Question 19

A block of metal is lying on the floor of a bus. The maximum acceleration which can be given to the bus so that the block may remain at rest, will be-

Accepted Answer

μg. The friction ma needed must not exceed the maximum friction μmg; the largest acceleration is when ma = μmg, giving a = μg.

Question 20

A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction (μ) between the block and the wall is 0.5, the magnitude of frictional force acting on the block is (g = 9.8 m/s²)

Accepted Answer

0.98 N. Maximum friction = μN = 0.5 × 5 = 2.5 N, but the weight mg = 0.1 × 9.8 = 0.98 N is smaller, so friction only balances the weight: f = 0.98 N.

Friction — NEET Physics MCQs with Solutions

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