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AP Physics B : Forces

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65 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Motion In Two Dimensions

A 2kg box is at the top of a frictionless ramp at an angle of 60^o. 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.

If the angle of the ramp is decreased, which of the following statements is false?

Possible Answers:

The time to reach the ground will increase

The force due to gravity will decrease

The normal force will decrease

The net force on the box will decrease

Correct answer:

The normal force will decrease

Explanation:

When given a question about the angle of a ramp, compare it to the extreme angles: 0^o and 90^o.

1. When the ramp has an angle of 0^o, the net force 0. The force due to gravity must equal the normal force; thus the normal force is at a maximum value.

2. When the angle of the ramp is 90^o, the full force of gravity is experienced by the box, and there is no normal force. The net force is equal to the force of gravity. Remember that $F_g=mgsin\Theta$ , so then theta is 90^o, force of gravity is at a maximum.

When we decrease the angle of the ramp, we get closer to scenario 1. As a result, we can conclude that the normal force on the box increases, rather than decreases.

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Example Question #1 : Understanding Normal Force

A woman is standing on a scale in an elevator as it accelerates upward. The elevator then stops accelerating and continues upward at a constant speed. Which of the following statements is true?

Possible Answers:

According to the scale, the woman weighed more when the elevator was accelerating

According to the scale, the woman's weight did not change throughout the motion

According to the scale, the woman weighed less when the elevator was accelerating

According to the scale, the woman weighed more when the elevator was moving at a constant speed

We must know the woman's mass to draw any conclusions

Correct answer:

According to the scale, the woman weighed more when the elevator was accelerating

Explanation:

The normal force of the woman is measured by the scale. While accelerating upward, the scale should read a larger weight than when it is at rest. When the elevator is moving upward at a constant speed, the scale should read the same as when it is at rest. This is because the normal force is generated to counter the downward forces pushing against the floor. When the elevator is accelerating, there is a net upward force from the acceleration as well as the normal force to counter gravity. The normal force generates an upward acceleration. When moving at a constant speed, there is no upward acceleration and the normal force acts only to counter gravity. The normal force, and scale reading, will thus be greater during the period of acceleration.

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Example Question #1 : Understanding Normal Force

A 10kg block is placed on a $30^\circ$ incline. What is the normal force of the incline on the box?

Possible Answers:

62.5N

78.4N

84.9N

98N

49N

Correct answer:

84.9N

Explanation:

To find the normal force on the incline, we use the relationship:

$F_{N}=mg\cos(\theta )$

This provides the magnitude of the force of gravity in the direction perpendicular to the incline. Normal force will always act in the direction perpendicular to the surface, and in this case will be equal and opposite to the force of gravity. We then plug in the mass and gravitational acceleration to find the normal force on this block:

$F_N=(10kg)(9.8\frac{m}{s^2})\cos(30^o)$

F_N=84.9N

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Example Question #1 : Understanding Normal Force

Five substitute players on a basketball team are sitting on the bench during a game. The bench weighs 400N and, altogether, the players weigh 4500N . Two players, weighing 1000N and 1200N , stand up. What is the difference in the normal force acting on the bench before and after these players leave?

Possible Answers:

2700N

2500N

4900N

2200N

1800N

Correct answer:

2200N

Explanation:

The normal force is generated as a result of a force against a solid surface. As per Newton's third law, the surface will exert an equal and opposite force on the object in contact. If an object is resting on a flat surface, then the normal force will be working to counter the weight of the object due to gravity.

The normal force acting on the bench with five players is equal and opposite to the total weight of the bench and players. Keep in mind that weight acts in the downward direction.

F_N=-F_g

F_g=-(400N+4500N)=-4900N

F_N=4900N

When the two players stand up, the new normal force is reduced.

F_g=(-4900N)-(-1000N-1200N)=-2700N

F_N=2700N

The difference in the normal force is:

4900N-2700N=2200N

We could also have found this change by adding the weights of the two players who stood.

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Example Question #1 : Forces

A person tries to lift a very heavy 70kg rock by applying an upward force of 500N , but is unable to move it upward. Calculate how much additional force was needed to lift the rock from the ground.

Possible Answers:

$143\ \text{N}$

$311\ \text{N}$

$362\ \text{N}$

$117\ \text{N}$

$186\ \text{N}$

Correct answer:

$186\ \text{N}$

Explanation:

First, calculate the gravitational force acting on the rock.

$F_g=mg=(70\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})=686\ \text{N}$

The exerts a force of 686N downward, meaning that if the person exerted at least 686N , then he or she would have been able to lift it up. Instead, the person applied only 500N . This means that the person needed to apply 186N of additional force to lift the rock.

686N-500N=186N

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Example Question #1 : Understanding Friction

A 20kg 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?

Possible Answers:

196N

216N

145.6N

245N

156.8N

Correct answer:

156.8N

Explanation:

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

F_N=-mg

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

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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Example Question #1 : Force Of Friction

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?

Possible Answers:

156.8N

6.3N

980.2N

16.4N

392.7N

Correct answer:

156.8N

Explanation:

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

$F_{friction}=\mu*F_{Normal}$

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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Example Question #1 : Force Of Friction

A man pulls a 50N box up a 30^o incline to rest at a height of . He exerts a total of 270J of work. What is the coefficient of friction on the incline?

Possible Answers:

0.82

0.35

We must know the mass of the box to solve

0.20

0.15

Correct answer:

0.20

Explanation:

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.

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

W_g=(50N)(4m)=200J

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

$W_{total}=W_g+W_f$

W_f=270J-200J=70J

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:

W=Fd

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.

W_f=F_f*d

70J=F_f(8m)

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

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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Example Question #1 : Understanding Friction

A 48kg 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 0.4 , what is the maximum acceleration that the truck can undergo without the crate slipping?

Possible Answers:

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

$a=8.56\frac{m}{s^2}$

$a=0.62\frac{m}{s^2}$

$a=12.97\frac{m}{s^2}$

$a=7.39\frac{m}{s^2}$

Correct answer:

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

Explanation:

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.

F_T=ma_T

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

F_T=F_f

$ma=\mu_smg$

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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Example Question #41 : Ap Physics B

A 2kg box is at the top of a ramp at an angle of 60^o. 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?

Possible Answers:

$\mu_k=0.08$

The kinetic coefficient of friction cannot be determined while the box is moving

$\mu_k=0.04$

$\mu_k=0.16$

Correct answer:

$\mu_k=0.08$

Explanation:

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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AP Physics B : Forces

Study concepts, example questions & explanations for AP Physics B

All AP Physics B Resources

Example Questions

Example Question #1 : Motion In Two Dimensions

Example Question #1 : Understanding Normal Force

Example Question #1 : Understanding Normal Force

Example Question #1 : Understanding Normal Force

Example Question #1 : Forces

Example Question #1 : Understanding Friction

Example Question #1 : Force Of Friction

Example Question #1 : Force Of Friction

Example Question #1 : Understanding Friction

Example Question #41 : Ap Physics B

All AP Physics B Resources

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