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AP Physics C: Mechanics : Mechanics Exam

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Example Questions

Example Question #91 : Ap Physics C

$\begin{align*}&\text{A block of mass }19kg\text{ rests on an incline of }69^{\circ}\\&\text{What is the frictional force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

186.39N

6.81N

174.01N

66.8N

Correct answer:

174.01N

Explanation:

Normal

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Example Question #92 : Ap Physics C

$\begin{align*}&\text{A block of mass }284kg\text{ rests on an incline of }48^{\circ}\\&\text{It is also partially supported by a rope.}\\&\text{The frictional force acting on the block is }207.04N\\&\text{What is the tension felt in the rope?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

2786.04N

2070.43N

1864.22N

2507.44N

Correct answer:

2507.44N

Explanation:

Normalwrope

$\begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{rope}=F_{mass}-\frac{F_{friction}}{sin(a)}\\&F_{rope}=2786.04N-\frac{207.04N}{sin(48^{\circ})}\\&F_{rope}=2507.44N\end{align*}$

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Example Question #93 : Ap Physics C

$\begin{align*}&\text{A block of mass }101kg\text{ rests on an incline of }68^{\circ}\\&\text{It is also partially supported by a rope with a tension of }693.57N\\&\text{What is the frictional force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

111.35N

275.6N

918.66N

371.16N

Correct answer:

275.6N

Explanation:

Normalwrope

$\begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{friction}=(101kg(9.81\frac{m}{s^2})-693.57N)(sin(68^{\circ}))\\&F_{friction}=275.6N\end{align*}$

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Example Question #15 : Friction And Normal Force

$\begin{align*}&\text{A block of mass }187kg\text{ rests on an incline of }70^{\circ}\\&\text{What is the frictional force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

627.43N

63.96N

1834.47N

1723.84N

Correct answer:

1723.84N

Explanation:

Normal

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Example Question #94 : Ap Physics C

$\begin{align*}&\text{A block of mass }208kg\text{ rests on an incline of }28^{\circ}\\&\text{What is the frictional force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}$

Possible Answers:

957.95N

1801.64N

183.65N

2040.48N

Correct answer:

957.95N

Explanation:

Normal

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Example Question #1 : Understanding Gravity

A block with a mass of 10kg is traveling at $50\frac{m}{s}$ when it impacts the ground. From how many meters off the ground was the block dropped?

Round to the nearest whole number.

Possible Answers:

124m

132m

126m

128m

130m

Correct answer:

128m

Explanation:

Set the gravitational potential energy and kinetic energy equal to each other and solve for the height.

$mgh = \frac{1}{2}mv^2$

Mass cancels.

$gh = \frac{1}{2}v^2$

Isolate the height and plug in our values.

$h = \frac{v^2}{2g} = \frac{(50\frac{m}{s})^2}{2*9.8\frac{m}{s^2}} = \frac{2500}{19.6} = 127.551m$

Rounding this gives h = 128m .

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Example Question #92 : Ap Physics C

An object of mass $\small m$ is dropped from a tower. The object's drag force is given by $\small F_{d}=-bv^{2}$ where $\small b$ is a positive constant. What will the objects terminal velocity be?

Possible Answers:

$v_{term}=\sqrt{\frac{mg}{b}}$

$\small v_{term}=\sqrt{\frac{g}{b}}$

$\small \small v_{term}=\frac{mg}{b}$

$\small \small v_{term}=\frac{g}{b}$

$\small \small v_{term}=\sqrt{\frac{b}{mg}}$

Correct answer:

$v_{term}=\sqrt{\frac{mg}{b}}$

Explanation:

To find terminal velocity, set the magnitude of the drag force equal to the magnitude of the force of gravity since when these forces are equal and opposite, the object will stop accelerating:

$\small F_{d}=mg$

$\small bv^{2}=mg$

Solve for

$v=\sqrt{\frac{mg}{b}}$

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Example Question #1 : Gravity

A ship of mass $\small m$ and an initial velocity of $\small v_0$ is coasting to a stop. The water exerts a drag force on the ship. The drag force is proportional to the velocity:

F_D =-kv where the negative sign indicates that the drag force acts in a direction opposite the motion. After the ship has coasted for a time equal to $\frac{2}{k}$ , how fast (in terms of $\small v_0$ ) will the ship be moving?

Possible Answers:

$\small \frac{v_0}{4}$

$\small 0.135 v_0$

$\small 0.63v_0$

$\small 0.37v_0$

$\small \frac{v_{0}}2{}$

Correct answer:

$\small 0.135 v_0$

Explanation:

The ship's equation of velocity (found by solving the first-order differential equation) is $\small v=v_0e^{-kt}$

Substitute $t=\frac{2}{k}$ and solve.

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Example Question #93 : Ap Physics C

A spherical asteroid has a hole drilled through the center as diagrammed below:

Hole through planet

Refer to the diagram above. An object that is much smaller than the asteroid is released from rest at the surface of the asteroid, at point a. How do the velocity and acceleration of the object compare at point b at the surface, and point a, located at the center of the asteroid?

Possible Answers:

$v_{a}<v_{b}$ and $a_{a}=a_{b}$

$v_{a}<v_{b}$ and $a_{a}>_{b}$

$v_{a}=v_{b}$ and $a_{a}=a_{b}$

$v_{a}< v_{b}$ and $a_{a}< a_{b}$

$v_{a}=v_{b}$ and $a_{a}< a_{b}$

Correct answer:

$v_{a}<v_{b}$ and $a_{a}>_{b}$

Explanation:

Because the gravitational force depends only on the mass beneath the object (which is the gravitational version of Gauss's Law for charge), the acceleration steadily decreases as the object falls, and drops to zero at the center. Nevertheless, the velocity keeps increasing as the object falls, it just does so more slowly.

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Example Question #91 : Mechanics Exam

With what minimum velocity must a rocket be launched from the surface of the moon in order to not fall back down due to the moon's gravity?

The mass of the moon is $7.2*10^{22}kg$ and its radius is 1.8*10^6m .

$\small G=6.67*10^{-11}\frac{m^2N}{kg^2}$

Possible Answers:

$730\frac{m}{s}$

$2.3\frac{km}{s}$

$1.6\frac{km}{s}$

$470\frac{m}{s}$

$520\frac{m}{s}$

Correct answer:

$2.3\frac{km}{s}$

Explanation:

Relevant equations:

$E_{total}=K+U_g$

$K = \frac{1}{2}mv^2$

$U_g=\frac{-GMm}{r}$

For the rocket to escape the moon's gravity, its minimum total energy is zero. If the total energy is zero, the rocket will have zero final velocity when it is infinitely far from the moon. If total energy is less than zero, the rocket will fall back to the moon's surface. If total energy is greater than zero, the rocket will have some final velocity when it is infinitely far away.

For the minimum energy case as the rocket leaves the surface:

$0 = K + U_g = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{r}$

Rearrange energy equation to isolate the velocity term.

$v_{esc}^2 = \frac{GMm}{\frac{1}{2}mr}$

$v_{esc}=\sqrt{\frac{2GM}{r}}$

Substitute in the given values to solve for the velocity.

$v_{esc}=\sqrt{\frac{2(6.67*10^{-11}\frac{m^2N}{kg^2})(7.2*10^{22}kg)}{(1.8*10^6m)}}=2310\frac{m}{s}\approx 2.3\frac{km}{s}$

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AP Physics C: Mechanics : Mechanics Exam

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #91 : Ap Physics C

Example Question #92 : Ap Physics C

Example Question #93 : Ap Physics C

Example Question #15 : Friction And Normal Force

Example Question #94 : Ap Physics C

Example Question #1 : Understanding Gravity

Example Question #92 : Ap Physics C

Example Question #1 : Gravity

Example Question #93 : Ap Physics C

Example Question #91 : Mechanics Exam

All AP Physics C: Mechanics Resources

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