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AP Physics 1: Algebra-Based Practice Quiz: Torque and Work

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

According to the provided principles, under which of the following conditions does a torque do work on a rigid system?

All Questions (10)

According to the provided principles, under which of the following conditions does a torque do work on a rigid system?

A) When the torque is exerted over an angular displacement.

B) When the torque is constant, regardless of displacement.

C) When the object's angular velocity is zero.

D) Whenever a net torque is applied to the system.

Correct Answer: A

The content states, "A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement." This means work is done only when there is both a torque and an angular displacement.

A constant torque of 10 N·m is applied to a flywheel, causing it to rotate through an angular displacement of 2.0 radians. How much work is done on the flywheel by the torque?

A) 5.0 J

B) 10 J

C) 12 J

D) 20 J

Correct Answer: D

The work done by a constant torque is calculated using the formula W = τΔθ. Given τ = 10 N·m and Δθ = 2.0 rad, the work done is W = (10 N·m)(2.0 rad) = 20 J.

A graph of torque as a function of angular position is created for a rigid system. The area under the curve between two angular positions, θ₁ and θ₂, represents which of the following quantities?

A) The change in angular momentum of the system.

B) The average torque exerted on the system.

C) The work done on the system by the torque.

D) The rotational inertia of the system.

Correct Answer: C

The provided content explicitly states, "Work done on a rigid system by a given torque can be found from the area under the curve of a graph of torque as a function of angular position."

When a net positive torque does positive work on a rigid system that is free to rotate, what is the direct consequence for the system?

A) The system's rotational inertia decreases.

B) The system's angular position remains constant.

C) Energy is transferred into the system.

D) Energy is transferred out of the system.

Correct Answer: C

The content states, "A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement." Positive work corresponds to energy being transferred into the system, typically increasing its rotational kinetic energy.

The torque on a rotating object varies linearly with its angular position, starting from 0 N·m at θ = 0 rad and increasing to 8 N·m at θ = 4 rad. What is the total work done by the torque over this interval?

A) 8 J

B) 16 J

C) 32 J

D) 64 J

Correct Answer: B

The work done is the area under the torque vs. angular position graph. In this case, the graph forms a triangle with a base of Δθ = 4 rad and a height of τ_max = 8 N·m. The area of a triangle is (1/2) * base * height. Therefore, W = (1/2) * (4 rad) * (8 N·m) = 16 J.

The equation W = τΔθ is used to calculate the work done by a torque on a rigid system. For this equation to be valid in this specific form, what must be true about the torque, τ?

A) The torque must be constant.

B) The torque must be zero.

C) The torque must be increasing linearly.

D) The torque must be perpendicular to the angular displacement.

Correct Answer: A

The formula W = τΔθ is a simplified case for a constant torque. If the torque varies, one must find the area under the torque vs. angular position graph, as stated in the provided content.

A constant torque τ does work W on a disk as it rotates through an angle Δθ. If the same torque is applied but the disk rotates through an angle of 3Δθ, what is the new work done on the disk?

A) W/3

B) W

C) 3W

D) 9W

Correct Answer: C

According to the formula W = τΔθ, work is directly proportional to the angular displacement, assuming the torque is constant. If the angular displacement is tripled, the work done is also tripled. The new work W' = τ(3Δθ) = 3(τΔθ) = 3W.

A mechanic applies a constant torque to a lug nut on a car wheel, but the nut does not rotate. How much work is done on the lug nut by the mechanic's torque?

A) A positive amount, because a torque is applied.

B) A negative amount, because the nut resists the torque.

C) Zero, because there is no angular displacement.

D) It cannot be determined without knowing the magnitude of the torque.

Correct Answer: C

Work is done by a torque only when it is exerted over an angular displacement (W = τΔθ). If the lug nut does not rotate, the angular displacement Δθ is zero. Therefore, the work done is W = τ * 0 = 0 J.

A spinning top is slowing down due to a constant frictional torque. Over one full revolution, the frictional torque does work on the top. Which of the following statements correctly describes this work and energy transfer?

A) Positive work is done, transferring energy into the top.

B) Negative work is done, transferring energy out of the top.

C) Zero work is done, because the net displacement is zero after a full revolution.

D) The work done is equal to the top's rotational inertia.

Correct Answer: B

A frictional torque opposes the direction of rotation. When the torque and angular displacement are in opposite directions, the work done (W = τΔθ) is negative. According to the content, work transfers energy; negative work means energy is transferred out of the system, which is consistent with the top slowing down.

A graph shows the torque on a system as a function of its angular position. From θ = 0 to θ = π rad, the torque is constant at +5 N·m. From θ = π to θ = 2π rad, the torque is constant at -5 N·m. What is the net work done on the system over the total displacement from θ = 0 to θ = 2π rad?

A) 10π J

B) 5π J

C) 0 J

D) -5π J

Correct Answer: C

The total work is the net area under the torque-angular position graph. Work from 0 to π is W₁ = τ₁Δθ₁ = (+5 N·m)(π rad) = 5π J. Work from π to 2π is W₂ = τ₂Δθ₂ = (-5 N·m)(2π - π rad) = -5π J. The net work is W_net = W₁ + W₂ = 5π J - 5π J = 0 J.