AP Physics 1 : Work, Energy, and Power

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #1 : Newtonian Mechanics

A bodybuilder is in the midst of a an intense training session. He is currently bench pressing a bar with a mass of \displaystyle 250 kg. If he does six reps of this mass and his arms are \displaystyle 0.75 m 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:

\displaystyle 0 J

None of the other answers here

\displaystyle 510 J

\displaystyle 1225 J

\displaystyle 1875 J

Correct answer:

\displaystyle 0 J

Explanation:

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:

\displaystyle W_{net} = \Delta U+\Delta K

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 #1 : Newtonian Mechanics

A semi-truck carrying a trailer has a total mass of \displaystyle 1500 kg. If it is traveling up a slope of \displaystyle 5^{\circ} to the horizontal at a constant rate of \displaystyle 20\frac{m}{s}, how much power is the truck exerting?

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

Possible Answers:

\displaystyle 34000W

\displaystyle 42000W

\displaystyle 21000W

\displaystyle 12000W

\displaystyle 26000W

Correct answer:

\displaystyle 26000W

Explanation:

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:

\displaystyle P = \frac{\Delta U}{t}=\frac{mg\Delta h}{t}

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

\displaystyle h = d\cdot sin (\theta)

\displaystyle P = \frac{mgd\cdot sin(\theta)}{t}

We can substitute velocity into this equation:

\displaystyle P = mgv\cdot sin(\theta)

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

\displaystyle P = (1500kg)(10\frac{m}{s^2})(20\frac{m}{s})\left sin(5^{\circ})\right = 26000W

Example Question #1 : Work, Energy, And Power

An upward force is applied to lift a \displaystyle 20kg bag a to a height of \displaystyle 5m. The bag is lifted at a constant speed. What is the work done on the bag?

Possible Answers:

\displaystyle 1000J

\displaystyle 500J

\displaystyle 1500J

\displaystyle 450J

\displaystyle 100J

Correct answer:

\displaystyle 1000J

Explanation:

Work is change in energy, so at a height of \displaystyle 5m, the bag has more potential energy that when on the ground (zero gravitational potential energy). potential energy is equal to:

\displaystyle mgh=20kg\cdot 10\frac{m}{s^2}\cdot 5m = 1000J

Alternatively, \displaystyle Work = Fd\cos \Theta.

\displaystyle \Theta is zero in this case, and \displaystyle F=mg=20kg\cdot 10\frac{m}{s^2}=200N

\displaystyle Work = 200N\cdot 5m = 1000J

Example Question #1 : Newtonian Mechanics

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:

\displaystyle 286.8kJ

\displaystyle 819.2kJ

\displaystyle 409.6kJ

\displaystyle 573.6kJ

\displaystyle 1000kJ

Correct answer:

\displaystyle 573.6kJ

Explanation:

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:

\displaystyle W=F\cdot d=Fdsin(\theta)

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

\displaystyle W=200N(5km)sin(35^{\circ})

\displaystyle W=573.6kJ

Example Question #2 : Newtonian Mechanics

Determine the work done by nonconservative forces if an object with mass 10kg is shot up in the air at \displaystyle 30\frac{m}{s} returns to the same height with speed \displaystyle 27\frac{m}{s}.

Possible Answers:

\displaystyle 171J

\displaystyle -171J

\displaystyle 1710J

\displaystyle -1710J

Correct answer:

\displaystyle -1710J

Explanation:

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

\displaystyle W_{outside}=\Delta PE+\Delta KE

 Here, \displaystyle \Delta PE is the change in potential energy, and \displaystyle \Delta KE is a change in kinetic energy. \displaystyle \Delta PE is \displaystyle 0 because the object returns to the same height as when it was launched. \displaystyle \Delta KE however has changed because the object's velocity has changed.  Recall that the formula for the change in kinetic energy is given by:

\displaystyle \Delta KE= mv^2_{final}-mv^2_{initial}

Here \displaystyle m is the mass of the object, \displaystyle v_{final} is the final velocity of the object and \displaystyle v_{initial} is the initial velocity of the object. 

In our case:

\displaystyle m=10kg\displaystyle v_{final}=45 \frac{m}{s}, and \displaystyle v_{initial}=50 \frac{m}{s}

\displaystyle \Delta KE= mv^2_{final}-mv^2_{initial}

\displaystyle \Delta KE=10kg*\left(27\frac{m}{s}\right)^2-10kg*\left(30\frac{m}{s}\right)^2

\displaystyle \Delta KE= 7290J-9000J= -1710J

\displaystyle W_{outside}=\Delta PE+\Delta KE

\displaystyle W_{outside}=-1710J 

Example Question #2 : Work, Energy, And Power

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

Possible Answers:

equal to the force exerted 

zero

equal to the kinetic energy of the object

equal to the force exerted multiplied by the displacement

Correct answer:

zero

Explanation:

Recall that work can be defined as:

\displaystyle W=F*d*cos(\theta)

Here, \displaystyle F is the magnitude of the force vector, \displaystyle d is the magnitude of the displacement vector, and \displaystyle \theta 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 \displaystyle 90^o. Since \displaystyle cos(90^o)=0, work done by the centripetal force on an object moving in a circle is always \displaystyle 0.

Example Question #2 : Newtonian Mechanics

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:

\displaystyle 3000kJ

\displaystyle 3000J

\displaystyle 6000kJ

\displaystyle 1500kJ

\displaystyle 1500J

Correct answer:

\displaystyle 3000kJ

Explanation:

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:

\displaystyle Work=3\ km(1000N)

\displaystyle Work=3000\ kJ

Example Question #3 : Newtonian Mechanics

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 \displaystyle 700N for \displaystyle 15 sec. She then pulls on the phone with a force of \displaystyle 900N for \displaystyle 10 sec. 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:

\displaystyle 0J

\displaystyle 900J

\displaystyle 136.67J

\displaystyle 19,500J

\displaystyle 1600J

Correct answer:

\displaystyle 0J

Explanation:

Work is a measure of force and displacement \displaystyle (W = F \cdot s). Because C.J. did not move the phone at all, no work was done.

Example Question #1 : Work, Energy, And Power

Fido, a small dog that weighs \displaystyle 100. N, sees a bird in a tree and climbs straight up, at constant velocity, with an average power of \displaystyle 640. W. If it takes Fido \displaystyle 8.50 sec to reach the branch upon which the bird is perched, how high is that branch?

Possible Answers:

\displaystyle 109m

\displaystyle 54.4m

\displaystyle 63.0m

\displaystyle 68.1m

\displaystyle 10.9m

Correct answer:

\displaystyle 54.4m

Explanation:

We can assume that the dog must carry his entire weight up the tree, and therefore a \displaystyle 100N force is exerted. Using the equation

 \displaystyle Power = \frac{Force(distance)}{time} 

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

\displaystyle 640\:W= \frac{100N(distance)}{8.5 \:seconds}

\displaystyle distance = 54.4m

Example Question #1 : Work, Energy, And Power

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

Possible Answers:

\displaystyle 500J

\displaystyle 1000J

\displaystyle 0J

\displaystyle 50J

Correct answer:

\displaystyle 0J

Explanation:

The answer is \displaystyle 0J 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:

\displaystyle W=f sin{\theta}d

Here, \displaystyle \theta is zero, so \displaystyle sin{\theta} is also zero, making the entire term, and thus the work zero.

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