Forces - AP Physics B

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Question

A astronaut on Earth uses a space elevator to get to a space station. From the ground, the elevator accelerates upwards at $20\frac{m}{s^2}$ . During this acceleration, what is the normal force acting on the astronaut?

Answer

This question essentially asks how much the astronaut will weigh during acceleration. The two forces acting on the astronaut are the downward gravitational force and the upward normal force. The net force acting on this astronaut is .

Notice that the net force is set to equal to to show that the astronaut is accelerating due to the elevator.

Isolate the normal force.

Use the given values to solve.

$F_N=(72kg)(20\frac{m}{s^2}+9.8\frac{m}{s^2})$

$F_N=2145.6\ \text{N}$

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Question

A astronaut on Earth uses a space elevator to get to a space station. From the ground, the elevator accelerates upwards at $20\frac{m}{s^2}$ . During this acceleration, what is the normal force acting on the astronaut?

Answer

This question essentially asks how much the astronaut will weigh during acceleration. The two forces acting on the astronaut are the downward gravitational force and the upward normal force. The net force acting on this astronaut is .

Notice that the net force is set to equal to to show that the astronaut is accelerating due to the elevator.

Isolate the normal force.

Use the given values to solve.

$F_N=(72kg)(20\frac{m}{s^2}+9.8\frac{m}{s^2})$

$F_N=2145.6\ \text{N}$

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Question

A 2kg box is at the top of a ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

Imagine that the net force on the box is 16.5N when sliding down the ramp. What is the coefficient of kinetic friction for the box?

Answer

Since the box is moving when the net force on the box is determined, we can calculate the coefficient of kinetic friction for the box. The first step is determining what the net force on the box would be in the absence of friction. The net force on the box is given by the equation $\small F = mgsin\Theta$ .

$\small F = (2kg)(10\frac{m}{s^{2}})sin(60^{\circ}) = 17.3N.$

The difference between the frictionless net force and the net force with friction is 0.8N. This means that the force of kinetic friction on the box is 0.8N, acting opposite the direction of motion. Knowing this, we can solve for the coefficient of kinetic friction using the equation $\small f_{k} = \mu_{k}F_{n}$

$\small 0.8 = (mgcos\Theta)(\mu _{k})$

$\small 0.8 = (2kg)(10cos(60^{\circ}))\mu_{k}$

$\small \mu_{k} = 0.08$

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Question

A crate is loaded onto a pick-up truck, and the truck speeds away without the crate sliding. If the coefficient of static friction between the truck and the crate is , what is the maximum acceleration that the truck can undergo without the crate slipping?

Answer

In order for the crate to not slide, the truck has to exert a frictional force on it. The force of friction is related to the normal force by the coefficient of friction.

$F_{friction}=\mu_sF_N=\mu_smg$

This frictional force comes from the acceleration of the truck, based on Newton's second law.

The two forces will be equal when the truck is at maximum acceleration without the crate moving.

$ma=\mu_smg$

Solve for the acceleration.

$a=\mu_sg$

$a=(0.4)(9.8\frac{m}{s^2})$

$a=3.92\frac{m}{s^2}$

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Question

A $\small 40kg$ box is initially sitting at rest on a horizontal floor with a coefficient of static friction $\mu_s=0.4$ . A horizontal pushing force is applied to the box. What is the maximum pushing force that can be applied without moving the box?

Answer

The maximum force that can be applied will be equal to the maximum value of the static friction force. The formula for friction is:

$F_{friction}=\muF_{Normal}$

We also know that the normal force is equal and opposite the force of gravity.

$F_{Normal}=-(F_g)=-mg$

Substituting to the original equation, we can rewrite the force of friction.

$F_{Friction}=\mu-(mg)$

Using the given values for the coefficient of friction and mass, we can calculate the force using the acceleration of gravity.

$F_{friction}=0.4-(40 kg-9.8\frac{m}{s^2})= 156.8 N$

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Question

A man pulls a box up a incline to rest at a height of . He exerts a total of of work. What is the coefficient of friction on the incline?

Answer

Work is equal to the change in energy of the system. We are given the weight of the box and the vertical displacement, which will allow us to calculate the change in potential energy. This will be the total work required to move the box against gravity.

$W_g=\Delta PE=mg\Delta h$

The remaining work that the man exerts must have been used to counter the force of friction acting against his motion.

$W_{total}=W_g+W_f$

Now we know the work performed by friction. Using this value, we can work to solve for the force of friction and the coefficient of friction. First, we will need to use a second formula for work:

In this case, the distance will be the distance traveled along the surface of the incline. We can solve for this distance using trigonometry.

$\sin(\theta)=\frac{opp}{hyp}$

$\sin(30^o)=\frac{4m}{d}=$

$d=\frac{4m}{\frac{1}{2}}=8m$

We know the work done by friction and the distance traveled along the incline, allowing us to solve for the force of friction.

$F_f=\frac{70J}{8m}=8.75N$

Finally, use the formula for frictional force to solve for the coefficient of friction. Keep in mind that the force on the box due to gravity will be equal to $mg\cos(\theta)$ .

$F_f=\mu F_N=\mu(-F_g)$

$F_g=mg\cos(\theta)$

$F_f=\mu(-mg\cos(\theta))$

Plug in our final values and solve for the coefficient of friction.

$8.75N=-\mu(-50N)\cos(30^o)=\mu(50N)(0.866)$

$\mu=\frac{8.75N}{50N(0.866)}=0.20$

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Question

A crate is sitting on a rough floor with coefficient of static friction $\mu=0.8$ . A man tries to push the crate horizontally. What force must the man exert on the crate in order to start moving the crate?

Answer

We know that normal force on a flat surface is equal and opposite the force of gravity.

$F_N=-(20kg)(-9.8\frac{m}{s^2})=196N$

Then, find the friction force between the crate and the floor using the equation:

$F_f=\mu F_N$

$F_{f}=(196N)(0.8)=156.8N$

This force is the minimum force required to start moving the crate.

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Question

A 2kg box is at the top of a ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

Imagine that the net force on the box is 16.5N when sliding down the ramp. What is the coefficient of kinetic friction for the box?

Answer

Since the box is moving when the net force on the box is determined, we can calculate the coefficient of kinetic friction for the box. The first step is determining what the net force on the box would be in the absence of friction. The net force on the box is given by the equation $\small F = mgsin\Theta$ .

$\small F = (2kg)(10\frac{m}{s^{2}})sin(60^{\circ}) = 17.3N.$

The difference between the frictionless net force and the net force with friction is 0.8N. This means that the force of kinetic friction on the box is 0.8N, acting opposite the direction of motion. Knowing this, we can solve for the coefficient of kinetic friction using the equation $\small f_{k} = \mu_{k}F_{n}$

$\small 0.8 = (mgcos\Theta)(\mu _{k})$

$\small 0.8 = (2kg)(10cos(60^{\circ}))\mu_{k}$

$\small \mu_{k} = 0.08$

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Question

A crate is loaded onto a pick-up truck, and the truck speeds away without the crate sliding. If the coefficient of static friction between the truck and the crate is , what is the maximum acceleration that the truck can undergo without the crate slipping?

Answer

In order for the crate to not slide, the truck has to exert a frictional force on it. The force of friction is related to the normal force by the coefficient of friction.

$F_{friction}=\mu_sF_N=\mu_smg$

This frictional force comes from the acceleration of the truck, based on Newton's second law.

The two forces will be equal when the truck is at maximum acceleration without the crate moving.

$ma=\mu_smg$

Solve for the acceleration.

$a=\mu_sg$

$a=(0.4)(9.8\frac{m}{s^2})$

$a=3.92\frac{m}{s^2}$

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Question

A $\small 40kg$ box is initially sitting at rest on a horizontal floor with a coefficient of static friction $\mu_s=0.4$ . A horizontal pushing force is applied to the box. What is the maximum pushing force that can be applied without moving the box?

Answer

The maximum force that can be applied will be equal to the maximum value of the static friction force. The formula for friction is:

$F_{friction}=\muF_{Normal}$

We also know that the normal force is equal and opposite the force of gravity.

$F_{Normal}=-(F_g)=-mg$

Substituting to the original equation, we can rewrite the force of friction.

$F_{Friction}=\mu-(mg)$

Using the given values for the coefficient of friction and mass, we can calculate the force using the acceleration of gravity.

$F_{friction}=0.4-(40 kg-9.8\frac{m}{s^2})= 156.8 N$

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Question

A man pulls a box up a incline to rest at a height of . He exerts a total of of work. What is the coefficient of friction on the incline?

Answer

Work is equal to the change in energy of the system. We are given the weight of the box and the vertical displacement, which will allow us to calculate the change in potential energy. This will be the total work required to move the box against gravity.

$W_g=\Delta PE=mg\Delta h$

The remaining work that the man exerts must have been used to counter the force of friction acting against his motion.

$W_{total}=W_g+W_f$

Now we know the work performed by friction. Using this value, we can work to solve for the force of friction and the coefficient of friction. First, we will need to use a second formula for work:

In this case, the distance will be the distance traveled along the surface of the incline. We can solve for this distance using trigonometry.

$\sin(\theta)=\frac{opp}{hyp}$

$\sin(30^o)=\frac{4m}{d}=$

$d=\frac{4m}{\frac{1}{2}}=8m$

We know the work done by friction and the distance traveled along the incline, allowing us to solve for the force of friction.

$F_f=\frac{70J}{8m}=8.75N$

Finally, use the formula for frictional force to solve for the coefficient of friction. Keep in mind that the force on the box due to gravity will be equal to $mg\cos(\theta)$ .

$F_f=\mu F_N=\mu(-F_g)$

$F_g=mg\cos(\theta)$

$F_f=\mu(-mg\cos(\theta))$

Plug in our final values and solve for the coefficient of friction.

$8.75N=-\mu(-50N)\cos(30^o)=\mu(50N)(0.866)$

$\mu=\frac{8.75N}{50N(0.866)}=0.20$

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Question

A crate is sitting on a rough floor with coefficient of static friction $\mu=0.8$ . A man tries to push the crate horizontally. What force must the man exert on the crate in order to start moving the crate?

Answer

We know that normal force on a flat surface is equal and opposite the force of gravity.

$F_N=-(20kg)(-9.8\frac{m}{s^2})=196N$

Then, find the friction force between the crate and the floor using the equation:

$F_f=\mu F_N$

$F_{f}=(196N)(0.8)=156.8N$

This force is the minimum force required to start moving the crate.

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Question

A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (ag) compare to the acceleration at the surface of Earth (g)?

Answer

The acceleration of gravity is given by the equation $a_{g} = \frac{GM}{r^{2}}$ , where G is constant.

For Earth, $a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$ .

For the new planet,

$a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$ .

So, the acceleration is the same in both cases.

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Question

What is the acceleration due to gravity on a planet on which an object with a mass of 20.0kg has a weight of 270N?

Answer

Solve the following equation for acceleration, using the values given in the question.

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

$a=\frac{270N}{20kg}=13.5\frac{m}{s^2}$

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Question

The moon's distance from the center of the Earth was decreased by a multiple of three. How would this affect the gravitational force of the Earth on the moon?

Answer

The law of gravitation is written as $\small F = \frac{Gm_{1}m_{2}}{r^{2}}$ , with G being equal to $\small \small 6.6710^{-11}\frac{m^3}{kgs^2}$ .

Since the radius of the two masses acting on each other is squared, and is found in the denominator, a decrease in the radius by a multiple of three will cause a nine-fold increase in the gravitational force.

$\small F_1 = \frac{Gm_{1}m_{2}}{r^{2}}\rightarrow F_2=\frac{Gm_{1}m_{2}}{(\frac{1}{3}r)^{2}}=9F_1$

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Question

A ball is thrown horizontally off a cliff of height of $\small 50m$ with an initial velocity of $\small 15\frac{m}{s}$ . How far from the cliff will the ball land?

Answer

First we will find the time required for the ball to reach the ground. Since the ball is thrown horizontally, it has no initial vertical component. We use the following equation to solve for the total flight time:

$\Delta y=v_{iy}t+\frac{1}{2}at^{2}$

We are given the change in height, initial velocity, and acceleration. Using these values, we can solve for the time. Note that the change in height will be negative, since the ball is traveling downward.

$-50m= (0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-50m=(-4.9\frac{m}{s^2})t^2$

$t=\sqrt{\frac{-50m}{-4.9\frac{m}{s^2}}}$

Finally, we use the horizontal velocity to find the distance traveled in . Remember that the horizontal velocity remains constant during projectile motion.

$v=\frac{d}{t}$

$d=vt=(15\frac{m}{s})(3.19s)=47.9m$

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Question

A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (ag) compare to the acceleration at the surface of Earth (g)?

Answer

The acceleration of gravity is given by the equation $a_{g} = \frac{GM}{r^{2}}$ , where G is constant.

For Earth, $a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$ .

For the new planet,

$a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$ .

So, the acceleration is the same in both cases.

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Question

What is the acceleration due to gravity on a planet on which an object with a mass of 20.0kg has a weight of 270N?

Answer

Solve the following equation for acceleration, using the values given in the question.

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

$a=\frac{270N}{20kg}=13.5\frac{m}{s^2}$

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Question

The moon's distance from the center of the Earth was decreased by a multiple of three. How would this affect the gravitational force of the Earth on the moon?

Answer

The law of gravitation is written as $\small F = \frac{Gm_{1}m_{2}}{r^{2}}$ , with G being equal to $\small \small 6.6710^{-11}\frac{m^3}{kgs^2}$ .

Since the radius of the two masses acting on each other is squared, and is found in the denominator, a decrease in the radius by a multiple of three will cause a nine-fold increase in the gravitational force.

$\small F_1 = \frac{Gm_{1}m_{2}}{r^{2}}\rightarrow F_2=\frac{Gm_{1}m_{2}}{(\frac{1}{3}r)^{2}}=9F_1$

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Question

A ball is thrown horizontally off a cliff of height of $\small 50m$ with an initial velocity of $\small 15\frac{m}{s}$ . How far from the cliff will the ball land?

Answer

First we will find the time required for the ball to reach the ground. Since the ball is thrown horizontally, it has no initial vertical component. We use the following equation to solve for the total flight time:

$\Delta y=v_{iy}t+\frac{1}{2}at^{2}$

We are given the change in height, initial velocity, and acceleration. Using these values, we can solve for the time. Note that the change in height will be negative, since the ball is traveling downward.

$-50m= (0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-50m=(-4.9\frac{m}{s^2})t^2$

$t=\sqrt{\frac{-50m}{-4.9\frac{m}{s^2}}}$

Finally, we use the horizontal velocity to find the distance traveled in . Remember that the horizontal velocity remains constant during projectile motion.

$v=\frac{d}{t}$

$d=vt=(15\frac{m}{s})(3.19s)=47.9m$

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