Laws of Motion Study Guide for Competitive Exams

Laws of Motion

Circular Motion Friction Newtons Laws

Q. A block is sliding down a frictionless incline of angle θ. If the incline has a coefficient of static friction μs, what is the maximum angle θ for which the block will not slide?

A. tan⁻¹(μs)
B. sin⁻¹(μs)
C. cos⁻¹(μs)
D. μs

Solution

The block will not slide if the component of gravitational force down the incline is less than or equal to the maximum static friction force, leading to θ = tan⁻¹(μs).

Correct Answer: A — tan⁻¹(μs)

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Q. A block is sliding down a frictionless incline. If the incline is now covered with a material that has a coefficient of kinetic friction of 0.3, how does this affect the acceleration of the block?

A. Increases acceleration
B. Decreases acceleration
C. No effect on acceleration
D. Acceleration becomes zero

Solution

The presence of kinetic friction opposes the motion, thus decreasing the acceleration of the block compared to a frictionless incline.

Correct Answer: B — Decreases acceleration

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Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of static friction is 0.5, what is the maximum static frictional force acting on the block?

A. 25 N
B. 50 N
C. 75 N
D. 100 N

Solution

Maximum static frictional force (Fs) = μs * N = μs * mg = 0.5 * 10 kg * 9.8 m/s² = 49 N, approximately 50 N.

Correct Answer: B — 50 N

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Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of kinetic friction is 0.3, what is the frictional force acting on the block when it is sliding?

A. 30 N
B. 20 N
C. 10 N
D. 15 N

Solution

Frictional force (f_k) = μ_k * N = μ_k * mg = 0.3 * 10 kg * 9.8 m/s² = 29.4 N, approximately 30 N.

Correct Answer: A — 30 N

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Q. A block of mass 5 kg is placed on a frictionless surface. A force of 10 N is applied to it. What is the acceleration of the block?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A block of mass 5 kg is placed on a frictionless surface. A force of 20 N is applied to it. What is the acceleration of the block?

A. 2 m/s²
B. 4 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A block of mass 5 kg is placed on a frictionless surface. If a force of 10 N is applied to it, what will be its acceleration?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 20 N is applied, what is the acceleration of the block? (Assume no friction)

A. 2 m/s²
B. 4 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Correct Answer: A — 2 m/s²

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Q. A block of mass 5 kg is resting on a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.3, what is the frictional force acting on the block when it is sliding?

A. 5 N
B. 10 N
C. 15 N
D. 20 N

Solution

Frictional force (Ff) = μk * N = μk * mg = 0.3 * (5 kg * 10 m/s²) = 15 N.

Correct Answer: B — 10 N

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Q. A block of mass 5 kg is resting on a horizontal surface. If the coefficient of static friction is 0.4, what is the maximum static frictional force acting on the block?

A. 10 N
B. 20 N
C. 15 N
D. 25 N

Solution

Maximum static frictional force (Fs) = μs * N = μs * mg = 0.4 * (5 kg * 10 m/s²) = 20 N.

Correct Answer: B — 20 N

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Q. A block slides down a frictionless incline of angle 30 degrees. If the incline has a coefficient of kinetic friction of 0.2, what is the acceleration of the block?

A. 4.9 m/s²
B. 3.9 m/s²
C. 2.9 m/s²
D. 1.9 m/s²

Solution

Net force = mg sin(30) - μmg cos(30). Acceleration a = (mg sin(30) - μmg cos(30))/m = g(sin(30) - μ cos(30)). Substituting g = 10 m/s² gives a = 10(0.5 - 0.2 * √3/2) = 4.9 m/s².

Correct Answer: B — 3.9 m/s²

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Q. A body is moving in a circular path of radius 4 m with a constant speed of 8 m/s. What is the net force acting on the body if its mass is 2 kg?

A. 4 N
B. 8 N
C. 16 N
D. 32 N

Solution

Centripetal force F = m(v²/r) = 2(8²/4) = 32 N.

Correct Answer: C — 16 N

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Q. A body is moving in a circular path with a radius of 10 m and completes one revolution in 5 seconds. What is its linear speed?

A. 2π m/s
B. 4π m/s
C. 10 m/s
D. 20 m/s

Solution

Linear speed (v) = Circumference / Time = (2π(10)) / 5 = 4π m/s.

Correct Answer: A — 2π m/s

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Q. A box is pushed across a floor with a force of 50 N. If the coefficient of kinetic friction is 0.4, what is the net force acting on the box if the normal force is 100 N?

A. 10 N
B. 20 N
C. 30 N
D. 40 N

Solution

Frictional force = μk * N = 0.4 * 100 N = 40 N. Net force = applied force - frictional force = 50 N - 40 N = 10 N.

Correct Answer: C — 30 N

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Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.3. If the normal force is 100 N, what is the net force acting on the box?

A. 20 N
B. 30 N
C. 50 N
D. 70 N

Solution

Frictional force = μk * N = 0.3 * 100 N = 30 N. Net force = applied force - frictional force = 50 N - 30 N = 20 N.

Correct Answer: B — 30 N

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Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.4. What is the acceleration of the box if its mass is 10 kg?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Net force = applied force - frictional force. Frictional force = μ_k * N = 0.4 * 10 kg * 9.8 m/s² = 39.2 N. Net force = 50 N - 39.2 N = 10.8 N. Acceleration = F/m = 10.8 N / 10 kg = 1.08 m/s², approximately 1 m/s².

Correct Answer: B — 2 m/s²

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Q. A car accelerates from rest at a rate of 2 m/s². What is the net force acting on the car if its mass is 1000 kg?

A. 200 N
B. 500 N
C. 1000 N
D. 2000 N

Solution

Using F = ma, F = 1000 kg * 2 m/s² = 2000 N.

Correct Answer: D — 2000 N

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Q. A car accelerates from rest to a speed of 20 m/s in 10 seconds. What is the distance covered by the car during this time?

A. 50 m
B. 100 m
C. 200 m
D. 400 m

Solution

Using the formula d = ut + 0.5at², where u = 0, a = (20 m/s) / 10 s = 2 m/s², we get d = 0 + 0.5 * 2 * (10)² = 100 m.

Correct Answer: B — 100 m

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Q. A car accelerates from rest to a speed of 30 m/s in 10 seconds. What is the distance covered by the car during this time?

A. 150 m
B. 300 m
C. 400 m
D. 600 m

Solution

Using the formula d = ut + 0.5at², where u = 0, a = (30 m/s) / 10 s = 3 m/s², we get d = 0 + 0.5 * 3 * (10)² = 150 m.

Correct Answer: B — 300 m

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Q. A car is moving in a circular path of radius 50 m with a constant speed of 20 m/s. What is the centripetal acceleration of the car?

A. 2 m/s²
B. 4 m/s²
C. 8 m/s²
D. 10 m/s²

Solution

Centripetal acceleration (a_c) = v²/r = (20 m/s)² / (50 m) = 8 m/s².

Correct Answer: B — 4 m/s²

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Q. A car is moving in a circular path of radius 50 m with a speed of 15 m/s. What is the angular displacement after 10 seconds?

A. 1 rad
B. 2 rad
C. 3 rad
D. 4 rad

Solution

Angular displacement θ = (v/r)t = (15/50)(10) = 3 rad.

Correct Answer: D — 4 rad

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Q. A car is moving in a circular track of radius 100 m at a speed of 20 m/s. What is the time period of one complete revolution?

A. 10 s
B. 20 s
C. 30 s
D. 40 s

Solution

Circumference = 2πr = 2π(100) = 200π m. Time period (T) = Circumference / Speed = 200π / 20 = 10π s.

Correct Answer: B — 20 s

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Q. A car is moving in a circular track of radius 100 m with a speed of 20 m/s. What is the time period of one complete revolution?

A. 10 s
B. 20 s
C. 30 s
D. 40 s

Solution

Circumference = 2πr = 2π(100) = 200π m. Time period (T) = Circumference / Speed = 200π / 20 = 10π s.

Correct Answer: B — 20 s

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Q. A car is moving in a circular track of radius 50 m with a speed of 15 m/s. What is the angular momentum of the car if its mass is 1000 kg? (2000)

A. 7500 kg m²/s
B. 10000 kg m²/s
C. 15000 kg m²/s
D. 20000 kg m²/s

Solution

Angular momentum L = mvr = 1000 * 15 * 50 = 750000 kg m²/s.

Correct Answer: A — 7500 kg m²/s

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Q. A car is moving in a circular track of radius 50 m with a speed of 15 m/s. What is the net force acting on the car if its mass is 1000 kg?

A. 200 N
B. 300 N
C. 400 N
D. 500 N

Solution

Centripetal force = mv²/r = 1000(15²)/50 = 4500 N.

Correct Answer: C — 400 N

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Q. A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Solution

The centripetal force required is provided by friction: F = mv^2/r. The frictional force is μmg. Setting them equal gives μ = v^2/(rg). Here, μ = (20^2)/(100*9.8) ≈ 0.4.

Correct Answer: C — 0.3

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Q. A car is negotiating a curve of radius 100 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent the car from skidding?

A. 0.15
B. 0.25
C. 0.30
D. 0.35

Solution

Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15 m/s)² / (100 m * 9.8 m/s²) ≈ 0.23.

Correct Answer: B — 0.25

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Q. A car of mass 1000 kg accelerates at 2 m/s². What is the net force acting on the car?

A. 2000 N
B. 500 N
C. 1000 N
D. 1500 N

Solution

Using F = ma, the net force is F = 1000 kg * 2 m/s² = 2000 N.

Correct Answer: A — 2000 N

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Q. A car of mass 1000 kg accelerates from rest to a speed of 20 m/s in 10 seconds. What is the net force acting on the car? (2000)

A. 100 N
B. 200 N
C. 300 N
D. 400 N

Solution

Acceleration a = (final speed - initial speed) / time = (20 m/s - 0) / 10 s = 2 m/s². Net force F = ma = 1000 kg * 2 m/s² = 2000 N.

Correct Answer: B — 200 N

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Q. A car of mass 1000 kg is moving at a speed of 15 m/s. What is the kinetic energy of the car?

A. 11250 J
B. 22500 J
C. 33750 J
D. 45000 J

Solution

Kinetic energy KE = (1/2)mv² = (1/2)(1000 kg)(15 m/s)² = 11250 J.

Correct Answer: B — 22500 J

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Laws of Motion

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