ap physics 1 review

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

• 0J

• 510J

• 1225J

• 1875J

Answer 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:

W_net=ΔU+ΔK

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.

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NGSS.HS-PS3-1

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

g=10m/s²

• 26000W

• 42000W

• 21000W

• 12000W

• 34000W

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NGSS.HS-PS2-1

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

• 100J

• 450J

• 500J

• 1000J

• 1500J

Tags

NGSS.HS-PS3-1

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 N. 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?

• 819.2kJ

• 1000kJ

• 286.8kJ

• 409.6kJ

• 573.6kJ

Tags

NGSS.HS-PS3-1

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

• equal to the kinetic energy of the object

• zero

• equal to the force exerted multiplied by the displacement

• equal to the force exerted 

Answer explanation

Recall that work can be defined as:

W=F∗d∗cos(θ)

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

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?

• 3000kJ

• 1500kJ

• 3000J

• 1500J

• 6000kJ

Answer 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:

Work=3 km(1000N)

Work=3000 kJ

Tags

NGSS.HS-PS3-1

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

• 19,500J

• 900J

• 136.67J

• 1600J

• 0J

Answer explanation

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

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

• 109m

• 54.4m

• 68.1m

• 10.9m

• 63.0m

Answer explanation

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

 Power=Force(distance)time 

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

640W=100N(distance)8.5seconds

distance=54.4m

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NGSS.HS-PS3-1

NGSS.HS-PS3-2

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

• 0J

• 1000J

• 500J

• 50J

Answer explanation

The answer is 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:

W=fsinθd

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

Tags

NGSS.HS-PS3-2

A model rocket weighing 5kg has a net propulsion force of 50N. Over a small period of time, the rocket speeds up (with constant acceleration) from an initial velocity of 20m/s to  a final velocity of 25m/s. Let us assume that the loss of mass due to fuel consumption is negligible and that the net force is along the direction of motion. How much net work was done on the rocket?

• 662.5J

• 562.5J

• 462.5J

• 262.5J

• 362.5J

Answer explanation

The net work done is equal to the change in kinetic energy. So we must find the kinetic energy at both the initial and final velocities and subtract.

KEi=.5mv2=(.5)(5kg)202=1000J

KEf=.5mv2=(.5)(5kg)252=1562.5J

W=ΔKE=1562.5−1000=562.5J

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NGSS.HS-PS2-1

NGSS.HS-PS3-1