### All AP Physics 1 Resources

## Example Questions

### Example Question #1 : Work

A bodybuilder is in the midst of a an intense training session. He is currently bench pressing a bar with a mass of . If he does six reps of this mass and his arms are long, how much work has been done on the bar between the time the bar was removed from its rack and placed back on the rack?

**Possible Answers:**

None of the other answers here

**Correct answer:**

The most important part of this question is noticing that it asks how much work has been done on the bar, not how much work the bodybuilder has exerted. Therefore we can use the work energy theorem:

Since the bar is initially at rest and returns to rest, the net work on the bar is zero. All of the energy exerted by the bodybuilder is counteracted by gravity.

Think about the system practically. Comparing the initial and final states, the bar is in the exact same position.

### Example Question #2 : Work

A semi-truck carrying a trailer has a total mass of . If it is traveling up a slope of to the horizontal at a constant rate of , how much power is the truck exerting?

**Possible Answers:**

**Correct answer:**

Since the truck is traveling at a constant rate, we know that all of the power exerted by the truck is going into a gain in potential energy. The power exerted will be a function of the change in potential energy over time. Therefore, we can write the following formula:

is a vertical height, so we need to write that as a function of distance traveled up the slope:

We can substitute velocity into this equation:

We have values for all of these variables, allowing us to solve:

### Example Question #3 : Work

An upward force is applied to lift a bag a to a height of . The bag is lifted at a constant speed. What is the work done on the bag?

**Possible Answers:**

**Correct answer:**

Work is change in energy, so at a height of , the bag has more potential energy that when on the ground (zero gravitational potential energy). potential energy is equal to:

Alternatively, .

is zero in this case, and

### Example Question #4 : Work

Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahead of her--five kilometers--and she's pulling with a force of 200 newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the wagon to get to the repair shop?

**Possible Answers:**

**Correct answer:**

Work exerted on an object is equal to the dot product of the force and displacement vectors, or the product of the magnitudes of the vectors and the sin of the angle between them:

The work exerted on the wagon in this problem is thus:

### Example Question #5 : Newtonian Mechanics

Determine the work done by nonconservative forces if an object with mass 10kg is shot up in the air at returns to the same height with speed .

**Possible Answers:**

**Correct answer:**

Since this question refers to work done by nonconservative forces, we know that:

Here, is the change in potential energy, and is a change in kinetic energy. is because the object returns to the same height as when it was launched. however has changed because the object's velocity has changed. Recall that the formula for the change in kinetic energy is given by:

Here is the mass of the object, is the final velocity of the object and is the initial velocity of the object.

In our case:

, , and

### Example Question #5 : Work

The work done by a centripetal force on an object moving in a circle at constant speed is __________.

**Possible Answers:**

equal to the kinetic energy of the object

equal to the force exerted

zero

equal to the force exerted multiplied by the displacement

**Correct answer:**

zero

Recall that work can be defined as:

Here, is the magnitude of the force vector, is the magnitude of the displacement vector, and is the angle between the directions of the force and displacement vectors. In the case of circular motion, the force vector is normal to the circle since it points inward, and the displacement vector is tangent to the circle. This means that the angle between the force vector and displacement is . Since , work done by the centripetal force on an object moving in a circle is always .

### Example Question #6 : Work

Raul is pushing a broken down car across the flat expanse of the Mojave to his shop.

If his shop is three kilometers away and he pushes with a Herculean force of one thousand newtons in the direction of his shop, how much work will be done on the car?

**Possible Answers:**

**Correct answer:**

Work is given by the dot product of force and displacement. Since both the force and displacement are in the same direction in this problem, work is simply the product of the two:

### Example Question #7 : Work

As a joke, Charlie glues C.J's phone to its receiver, which is bolted to her desk. Trying to extricate it, C.J. pulls on the phone with a force of for . She then pulls on the phone with a force of for . Unfortunately, all of her exertion is in vain, and neither the phone, nor receiver move at all. How much work did C.J. do on the phone in her 25 total seconds of pulling?

**Possible Answers:**

**Correct answer:**

Work is a measure of force and displacement . Because C.J. did not move the phone at all, no work was done.

### Example Question #8 : Work

Fido, a small dog that weighs , sees a bird in a tree and climbs straight up, at constant velocity, with an average power of . If it takes Fido to reach the branch upon which the bird is perched, how high is that branch?

**Possible Answers:**

**Correct answer:**

We can assume that the dog must carry his entire weight up the tree, and therefore a force is exerted. Using the equation

we can use the evidence provided by the problem to solve for distance.

### Example Question #9 : Work

A 50kg man pushes against a wall with a force of 100N for 10 seconds. How much work does the man accomplish?

**Possible Answers:**

**Correct answer:**

The answer is because no work is done. For work to be done a force must be exerted across a distance parallel to the direction of the force. In this case, a force is exerted by the man but the wall is stationary and since it does not move there is no distance for work to take place on. The formula for work can be written as:

Here, is zero, so is also zero, making the entire term, and thus the work zero.