Laws of Motion Study Guide for Competitive Exams

Laws of Motion

Circular Motion Friction Newtons Laws

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

A. 11250 J
B. 15000 J
C. 22500 J
D. 30000 J

Solution

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

Correct Answer: A — 11250 J

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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. If the brakes are applied and the car comes to a stop in 5 seconds, what is the average force applied by the brakes?

A. 2000 N
B. 4000 N
C. 5000 N
D. 6000 N

Solution

The change in momentum is 1000 kg * 20 m/s = 20000 kg·m/s. The average force is F = Δp/Δt = 20000 kg·m/s / 5 s = 4000 N.

Correct Answer: B — 4000 N

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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. What is the momentum of the car?

A. 2000 kg·m/s
B. 10000 kg·m/s
C. 5000 kg·m/s
D. 40000 kg·m/s

Solution

Momentum p = mv = 1000 kg * 20 m/s = 20000 kg·m/s.

Correct Answer: B — 10000 kg·m/s

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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. What is the net force required to bring it to rest in 5 seconds?

A. 4000 N
B. 2000 N
C. 1000 N
D. 500 N

Solution

First, find the deceleration: a = (final velocity - initial velocity) / time = (0 - 20 m/s) / 5 s = -4 m/s². Then, F = ma = 1000 kg * 4 m/s² = 4000 N.

Correct Answer: A — 4000 N

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Q. A car travels around a circular track of radius 50 m at a speed of 15 m/s. What is the centripetal force acting on the car if its mass is 1000 kg?

A. 450 N
B. 225 N
C. 150 N
D. 75 N

Solution

Centripetal force (F_c) = mv²/r = 1000 kg * (15 m/s)² / 50 m = 4500 N.

Correct Answer: A — 450 N

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Q. A conical pendulum consists of a mass attached to a string that swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. mg/cos(θ)
B. mg/sin(θ)
C. mg/tan(θ)
D. mg

Solution

Tension T = mg/cos(θ) to balance the vertical component of weight.

Correct Answer: A — mg/cos(θ)

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Q. A conical pendulum consists of a mass m attached to a string of length L, swinging in a horizontal circle. What is the expression for the tension in the string?

A. T = mg
B. T = mg/cos(θ)
C. T = mg/sin(θ)
D. T = m(v²/r)

Solution

In a conical pendulum, T = mg/cos(θ) where θ is the angle with the vertical.

Correct Answer: B — T = mg/cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical increases, what happens to the tension in the string?

A. Increases
B. Decreases
C. Remains the same
D. Becomes zero

Solution

As the angle increases, the vertical component of tension must increase to balance the weight.

Correct Answer: A — Increases

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. T = mg
B. T = mg/cos(θ)
C. T = mg/sin(θ)
D. T = mg tan(θ)

Solution

The vertical component of tension balances the weight: T cos(θ) = mg, thus T = mg/cos(θ).

Correct Answer: B — T = mg/cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension in the string and the gravitational force?

A. T = mg
B. T = mg/cos(θ)
C. T = mg/sin(θ)
D. T = mg/tan(θ)

Solution

Tension T provides the centripetal force and balances the weight, T = mg/cos(θ).

Correct Answer: B — T = mg/cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension in the string and the gravitational force acting on the pendulum bob?

A. T = mg
B. T = mg cos(θ)
C. T = mg sin(θ)
D. T = mg tan(θ)

Solution

Tension provides the vertical component to balance the weight: T cos(θ) = mg.

Correct Answer: B — T = mg cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension and the gravitational force acting on the pendulum bob?

A. T = mg
B. T = mg cos(θ)
C. T = mg sin(θ)
D. T = mg tan(θ)

Solution

The vertical component of tension balances the weight: T cos(θ) = mg.

Correct Answer: B — T = mg cos(θ)

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Q. A conical pendulum swings with a constant speed. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. mg/cos(θ)
B. mg/sin(θ)
C. mg/tan(θ)
D. mg

Solution

Tension T = mg/cos(θ) to balance the vertical component of weight.

Correct Answer: A — mg/cos(θ)

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Q. A cyclist is moving around a circular track of radius 100 m. If he completes one lap in 40 seconds, what is his average speed?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(100 m). Average speed = distance/time = (2π(100 m))/40 s ≈ 15.7 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is moving in a circular path of radius 10 m at a speed of 5 m/s. What is the angle of banking required to prevent slipping?

A. 30°
B. 45°
C. 60°
D. 90°

Solution

tan(θ) = v²/(rg) = (5²)/(10*9.8) = 0.25. θ = tan⁻¹(0.25) ≈ 14°.

Correct Answer: A — 30°

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Q. A cyclist is moving in a circular path of radius 15 m with a speed of 5 m/s. What is the angular velocity of the cyclist?

A. 0.2 rad/s
B. 0.5 rad/s
C. 1 rad/s
D. 2 rad/s

Solution

Angular velocity (ω) = v/r = 5 m/s / 15 m = 1/3 rad/s.

Correct Answer: B — 0.5 rad/s

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Q. A cyclist is moving in a circular path of radius 15 m with a speed of 6 m/s. What is the angular velocity of the cyclist?

A. 0.4 rad/s
B. 0.6 rad/s
C. 0.8 rad/s
D. 1.0 rad/s

Solution

Angular velocity (ω) = v/r = 6 m/s / 15 m = 0.4 rad/s.

Correct Answer: B — 0.6 rad/s

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Q. A cyclist is moving in a circular track of radius 30 m with a speed of 15 m/s. What is the net force acting on the cyclist if the mass of the cyclist is 60 kg?

A. 180 N
B. 120 N
C. 90 N
D. 60 N

Solution

Centripetal force F = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.

Correct Answer: A — 180 N

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Q. A cyclist is moving in a circular track of radius 30 m. If he completes one round in 12 seconds, what is his average speed?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(30) = 60π m. Average speed = distance/time = 60π m / 12 s = 5 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the angular velocity of the cyclist?

A. π/6 rad/s
B. π/3 rad/s
C. 2π/6 rad/s
D. 2π/3 rad/s

Solution

Angular velocity (ω) = 2π/T = 2π/12 = π/6 rad/s.

Correct Answer: B — π/3 rad/s

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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the average speed of the cyclist?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(30 m). Average speed = distance/time = (2π(30 m))/12 s ≈ 5 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist towards the center of the track?

A. 50 N
B. 75 N
C. 100 N
D. 125 N

Solution

Centripetal force (F_c) = mv²/r. Assuming mass m = 100 kg, F_c = (100 kg)(15 m/s)² / (30 m) = 75 N.

Correct Answer: C — 100 N

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Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist if the mass of the cyclist is 60 kg?

A. 180 N
B. 120 N
C. 90 N
D. 60 N

Solution

Centripetal force F_c = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.

Correct Answer: A — 180 N

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Q. A cyclist is negotiating a circular turn of radius 30 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent slipping?

A. 0.25
B. 0.5
C. 0.75
D. 1

Solution

Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15²)/(30*9.8) = 0.25.

Correct Answer: B — 0.5

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Q. A force of 10 N is applied to a 2 kg object at rest. What is the final velocity of the object after 5 seconds?

A. 2 m/s
B. 5 m/s
C. 10 m/s
D. 15 m/s

Solution

Using F = ma, a = F/m = 10 N / 2 kg = 5 m/s². Final velocity v = u + at = 0 + 5 * 5 = 25 m/s.

Correct Answer: B — 5 m/s

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Q. A force of 10 N is applied to a 2 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 4 N?

A. 6 N
B. 10 N
C. 14 N
D. 4 N

Solution

Net force = applied force - frictional force = 10 N - 4 N = 6 N.

Correct Answer: A — 6 N

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Q. A force of 10 N is applied to a 2 kg object. What is the object's acceleration?

A. 5 m/s²
B. 10 m/s²
C. 2 m/s²
D. 20 m/s²

Solution

Using F = ma, acceleration a = F/m = 10 N / 2 kg = 5 m/s².

Correct Answer: A — 5 m/s²

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Q. A force of 10 N is applied to a 2 kg object. What is the resulting acceleration?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, we have a = F/m = 10 N / 2 kg = 5 m/s².

Correct Answer: A — 2 m/s²

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Q. A force of 15 N is applied to a 3 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 5 N?

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

Solution

Net force = applied force - frictional force = 15 N - 5 N = 10 N.

Correct Answer: A — 10 N

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Q. A force of 15 N is applied to a 3 kg object. What is the object's acceleration?

A. 3 m/s²
B. 5 m/s²
C. 7 m/s²
D. 10 m/s²

Solution

Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².

Correct Answer: B — 5 m/s²

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Laws of Motion

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