### All AP Physics C: Mechanics Resources

## Example Questions

### Example Question #1 : Friction And Normal Force

A truck traveling at suddenly brakes and skids to a halt. If the coefficient of kinetic friction between the tires and the road is , how far does the truck skid before stopping?

**Possible Answers:**

**Correct answer:**

Relevant equations:

Determine the truck's change in kinetic energy. Note that the final kinetic energy will be zero because it has no final velocity.

Using the work-energy theorem, set change in kinetic energy equal to the work done by friction.

Substitute the equation for force of friction.

Use the given values for coefficient of friction, mass, and acceleration of gravity to solve for the distance.

### Example Question #1 : Friction And Normal Force

A force of 40N is applied on a 14kg box to move it to the right. If the coefficient of friction between the box and the floor is 0.2, what is the acceleration of the box?

**Possible Answers:**

**Correct answer:**

The two horizontal forces acting on the box are the applied force of 40N to the right, and the friction force ()to the left. So the net horizontal force is written as the equation below.

is the normal force. Solving for acceleration, a, we get the equation below.

We know the following from the question.

Thus, the acceleration is

### Example Question #1 : Forces

A crate is pulled across a horizontal floor at a constant velocity of . The coefficient of kinetic friction between the crate and the floor is . If the rope pulling the crate is angled at above the horizontal, what force must be applied to maintain this constant velocity?

**Possible Answers:**

**Correct answer:**

Relevant equations:

Write net force equations for horizontal and vertical directions:

Horizontal:

Vertical:

Since the crate is not accelerating, in each direction, implying that .

Horizontal:

Vertical:

Express the horizontal and vertical components of the pulling force in terms of the total pulling force, .

Horizontal:

Vertical:

Replace force of friction with its expression in terms of the normal force.

Horizontal:

Rearrange to isolate the normal force.

Horizontal:

Substitute this term for the normal force in the net vertical force equation.

Vertical:

Solve to isolate the pulling force.

Use the given values for the mass of the box, coefficient of friction, and acceleration of gravity to solve for the pulling force.

### Example Question #1 : Friction

A truck driver is making a delivery. He takes a 287kg box out of his semi-truck by having it slide down a 33^{o} ramp that is 5.5m long. To ensure that the box doesn't accelerate down the ramp, the driver pushes back on it so that the box's constant speed is . If the coefficient of friction is 0.4, calculate the work done by the frictional force.

**Possible Answers:**

**Correct answer:**

We can calculate the work done by friction by using .

It is just the product of the friction force and the length of the ramp. Rewrite work as .

Since this problem involves a box sliding down an incline, the normal force is .

Work is now written as the equation below.

We know the following information from the question.

using these values, we can calculate work from our equation.

### Example Question #1 : Friction

A string connected to a box sitting on the floor is used to apply a

force on the box as shown. The string is inclined at an angle of 30°

above the horizontal, the box has a mass of 15 kg, and the tension

in the string is 40 N. The box is at rest, but on the verge of slipping.

What is the coefficient of static friction between the box and the

floor?

**Possible Answers:**

**Correct answer:**

First, we know that the tension on the string is at an angle of degrees from the horizontal. We must find the x and y components of the tension force.

Next, we must find the normal force acting on the block. The normal force of the block itself is , but there is a string pulling on it. We must subtract the y component of tension to get the final normal force

Next, we know that the block is being pulled to the right, and the frictional force is acting in the opposite direction. So we get the equation:

### Example Question #11 : Friction And Normal Force

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**Correct answer:**

### Example Question #14 : Forces

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### Example Question #15 : Forces

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### Example Question #2 : Friction

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### Example Question #3 : Friction

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