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.

Mixed — NEET Physics MCQs with Solutions

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