Forces - AP Physics C: Electricity and Magnetism

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Question

A block with a mass of 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.

Answer

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 .

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Question

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

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?

Answer

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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Question

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?

Answer

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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Question

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:

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?

Answer

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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Question

A $\small 3kg$ paint bucket hangs from the right end of a meter stick, oriented horizontally. The left end of the meter stick rests on a fulcrum so that it may rotate about that point. A rope is to be attached $\small 60cm$ from the left end, so that the system does not rotate. What is the minimum force this rope must support, assuming the meter stick is massless?

Answer

Relevant equations:

$\tau = r \times F$

$\tau_{net} = I \alpha$

$\tau_{net}= \tau_{ccw}- \tau_{cw}$

Determine the clockwise torque caused by the bucket.

$\tau_{cw}=(1m)F_g = (1m)(3kg)(9.8\frac{m}{s^2})=29.4 Nm$

Write an expression for the counterclockwise torque caused by the rope.

$\tau_{ccw}=(0.60m)F_{rope}$

Combine the torque of the rope and the torque of the bucket into the net torque equation.

$\tau_{net}=(0.60m)F_{rope}-29.4Nm$

Since the system has no angular acceleration, net torque must be zero, allowing us to solve for the force of the rope.

$0 = (0.60m)F_{rope}-29.4Nm$

$F_{rope}=\frac{29.4Nm}{0.60m}$

$F_{rope}=49N$

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Question

An object at rest will remain at rest unless acted on by a(n) .

Answer

The correct answer is external force. External forces applied to an object will result in non-zero acceleration, causing the object to move. In contrast, internal forces contribute to the properties of the object and do not result in acceleration of the object.

Either positive or negative forces can result in the acceleration of an object. The sign on the force simply conveys information about its relative direction.

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Question

A satellite is in an elliptical orbit about the Earth. If the satellite needs to enter a circular orbit at the apogee of the ellipse, in what direction will it need to accelerate?

Answer

Since the speed of the satellite at apogee is too low for a circular orbit at that orbital radius, the satellite needs to speed up to circularize the orbit.

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Question

If $v_{escape}$ is the escape velocity from the surface of a planet of mass and radius . What is the velocity necessary for an object launched from the surface of a planet of the same mass, but with a radius that is , to escape the gravitational pull of this smaller, denser planet?

Answer

Escape velocity is the velocity at which the kinetic energy of an object in motion is equal in magnitude to the gravitational potential energy. This allows us to set two equations equal to each other: the equation for kinetic energy of an object in motion and the equation for gravitational potential energy. Therefore:

$\frac{1}{2}mv^{2}=\frac{GMm}{r}$

$mv^{2}=\frac{2GMm}{r}$

$v^{2}=\frac{2GM}{r}$

$v=\sqrt{\frac{2GM}{r}}$

In this case, $v=v_{escape}$ . Substitute.

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

Since the radius of the new planet is or $\frac{1}{4}$ the radius of the original planet, and they have the same masses, we can equate:

$v^{}=\sqrt{\frac{2GM}{\frac{R}{4}}}=\sqrt{4}\sqrt{\frac{2GM}{R}}=2\sqrt{\frac{2GM}{R}}=2v_{escape}$

Where $v_{escape}$ is the escape velocity of the larger planet and $v^{}$ is the escape velocity of the smaller planet.

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Question

A comet is in an elliptical orbit about the Sun, as diagrammed below:

The mass of the comet is very small compared to the mass of the Sun.

In the diagram above, how does the net torque on the comet due to the Sun's gravitational force compare at the two marked points?

Answer

Since the direction of the gravitational force is directly towards the center of the Sun, it lies in the plane of the comet's orbit. Since torque is $\small r\cdot F = Fr\sin \theta$ , the torque is always zero. That is why the comet's angular momentum is conserved in orbit.

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Question

A satellite of mass is in an elliptical orbit about the Earth. It's velocity at perigee is $\small v_{p}$ and its orbital radius at perigee is $\small r_{p}$ . If the radius at apogee is $\small 2r_{p}$ , what is its velocity at apogee?

Answer

Since the gravitational force cannot exert torque on the satellite about the Earth's center, angular momentum is conserved in this orbit:

$\small L_p=L_a$

$\small mv_pr_p=mv_ar_a$

$\small v_pr_p=v_a(2r_p)$

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Question

A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?

$g=10\frac{m}{s^2}$

Answer

When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.

In addition, the rocket experiences the downward force of its own weight. This is given by:

$W = mg = 1000 kg * 10 \frac{m}{s^2} = 10000 N$

We know that $F_{net} = ma$ .

We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.

We know that $F_{net} = F - W$ (since is directed upwards and is directed downwards).

$F_{net} = 30000 N - 10000 N = 20000 N$

Finally we solve for acceleration:

$a = \frac{F_{net}}{m}=\frac{20000N}{1000kg}=20\frac{m}{s^2}$

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Question

Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?

Answer

Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.

A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and as $T_{1}$ . Similarly, we will label the tension experienced by the rope between and as $T_{2}$ .

For the box on the right (of mass ) we know that the force pulls to the right and the rope pulls it to the left with a tension of $T_{1}$ .

We get the equation: $F - T_{1} = 2ma$ . is the mass of the box and is its acceleration (which is the same for the other boxes).

For the middle box (of mass ) we have a rope pulling on it on the right with tension $T_{1}$ and a rope pulling on the left with tension $T_{2}$ .

So we get the equation: $T_{1}- T_{2}= ma$ . is the mass of the box and is its acceleration (which is the same for the other boxes).

Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension $T_{2}$ .

So we get the equation: $T_{2}=3ma$ . is the mass of the box and is its acceleration (which is the same for the other boxes).

Now it is just a matter of simple substitution. We have three equations:

$T_{2}=3ma$

$T_{1}- T_{2}= ma$

$F - T_{1} = 2ma$

From them we can get an expression for the force. Isolate $T_{1}$ in the second equation.

$T_{1}= ma + T_{2}$

Substitute the expression of $T_{2}$ from the first equation and simplify.

Use this value of $T_{1}$ in the third equation to get an expression for the force.

$F = 2ma + T_{1} = 2ma + 4ma = 6ma$

So we have:

Solve for acceleration.

$a = \frac{F}{6m}$

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Question

A box is being pushed against a wall by a force as shown in the picture. If the coefficient of static friction between the box and the wall is $\mu$ , which of the following expressions represents what the force must be for the box not to fall?

Answer

This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force . The box experiences the downward force of its own weight given by . Finally, since the box is trying to slide down, friction opposes this motion. The following diagram shows these forces:

The key is that the box should NOT move.

For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.

Now, on the vertical axis we have fiction and the object's weight:

Finally, we use the equation for friction force to solve the problem.

$f= N\mu$

Substitute the weight, since it is equal to the force of friction.

$N\mu=mg$

Isolate the normal force.

$N=\frac{mg}{\mu }$

Since the normal force is equal to the applied force, this is our final expression.

$F = N = \frac{mg}{\mu }$

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Question

A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?

Answer

You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.

We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.

$\cos\theta=\frac{adj}{hyp}\rightarrow adj=(hyp)(\cos\theta)$

We can solve for the horizontal component of the force using the applied force as the hypotenuse:

$F_{x} = Fcos(60º) = 20N*\frac{1}{2} = 10N$

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Question

A book of mass 1kg is held to a vertical wall by a person's hand applying a 20N force directly toward the wall. The wall has a static friction coefficient of 0.3 and a kinetic friction coefficient of 0.2. With the book held at rest, what the is frictional force keeping the book from sliding down the wall?

Answer

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

$F_f=\mu_sF_N$

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

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Question

Weightlessness is experienced when the normal force equals .

Answer

"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

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Question

$\begin{align}&\text{A block of mass }190kg\text{ rests on an incline of }33^{\circ}\&\text{It is also partially supported by a rope with a tension of }559.17N\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), the force of the rope supporting it, a normal force,}\&\text{and since it’s on an incline, a frictional force to balance}\&\text{it all out. Notice that the tension in the rope mitigates some}\&\text{of the gravitational force (as illustrated in the upper left}\&\text{corner) to lessen what is applied to the slope:}\&F_{mass}=mg\&F_{normal}=(F_{mass}-F_{rope})cos(a)\&F_{friction}=(F_{mass}-F_{rope})sin(a)\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\&F_{normal}=1094.24N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }233kg\text{ rests on an incline of }81^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=233kg(9.81\frac{m}{s^2})(cos(81^{\circ}))\&F_{normal}=357.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }187kg\text{ rests on an incline of }83^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=187kg(9.81\frac{m}{s^2})(cos(83^{\circ}))\&F_{normal}=223.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }267kg\text{ rests on an incline of }49^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=267kg(9.81\frac{m}{s^2})(cos(49^{\circ}))\&F_{normal}=1718.4N\end{align*}$

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