AP Physics C: Mechanics Practice Quiz: Torque and Work

Correct Answer: A

For a constant torque, the integral $W=\int \tau d\theta$ simplifies to $W = \tau \Delta\theta$, which is the product of the torque's magnitude and the angular displacement.

Correct Answer: B

The integral sums up the infinitesimal amounts of work ($dW = \tau d\theta$) done as the system rotates from an initial angular position $\theta_{1}$ to a final angular position $\theta_{2}$. This represents the total accumulation of the torque's effect over the angular displacement.

Correct Answer: D

The work done is the area under the curve of a torque vs. angular position graph. For a constant torque, this area is a rectangle. Work = Torque × Angular Displacement = (10 N·m) × (4 rad) = 40 J.

Correct Answer: B

As stated in the content, the 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. This is the graphical representation of the integral $W=\int \tau d\theta$.

Correct Answer: C

The work done is the integral of torque with respect to angular position: $W=\int_{0}^{3} 2\theta d\theta$. Evaluating the integral gives $[\theta^2]_{0}^{3} = 3^2 - 0^2 = 9$ J. Alternatively, this can be seen as the area of a triangle on a torque vs. angular position graph with base 3 rad and height $\tau(3) = 2(3) = 6$ N·m. Area = (1/2) * base * height = (1/2) * 3 * 6 = 9 J.

Correct Answer: C

Work done by a torque is calculated as the integral of torque over an angular displacement, $W=\int \tau d\theta$. If the system does not rotate, the angular displacement (d$\theta$) is zero, and therefore the work done is zero, regardless of the magnitude of the applied torque.

Correct Answer: C

The work done by a collection of torques is the sum of the work done by each individual torque. The net torque is $\tau_{net} = \tau_1 + \tau_2$. The total work is $W_{net} = \tau_{net} \Delta\theta = (\tau_1 + \tau_2) \Delta\theta = \tau_1 \Delta\theta + \tau_2 \Delta\theta$.

Correct Answer: B

The torque is constant, so work is the product of the torque and the angular displacement. The angular displacement is $\Delta\theta = \theta_{final} - \theta_{initial} = 12 - 10 = 2$ rad. The work done is $W = \tau \Delta\theta = (-5 \text{ N·m})(2 \text{ rad}) = -10$ J. The negative sign indicates that the torque opposes the displacement.

Correct Answer: B

For a constant torque, work is directly proportional to the angular displacement ($W = \tau \Delta\theta$). If the angular displacement is halved, the work done will also be halved. Therefore, the new work done is 100 J / 2 = 50 J.