If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# AP Physics 1 review of Energy and Work

In this video David explains the concepts in Work and Energy and does an example problem for each concept. Link for document: https://www.dropbox.com/s/t1w6xlnkozzel17/Energy%20review.pdf?dl=0. Created by David SantoPietro.

## Want to join the conversation?

• at , does the direction of the velocities not matter because the squares will make them positive anyways? • When a force is perpendicular to the direction of motion, why is there 0 work being done on the object? • Think of an object already moving horizontally across the floor as a constant speed.
There is a string attached to the object, which you can pull straight up on.
If you pull straight up on the string with insufficient force to lift it, it is not moved. Since there is no displacement, there can be no work done (something times zero is always zero).
If you pull straight up on the string with MORE than enough force to lift it, the object moved off the ground and is given a vertical displacement. This gives something for the force to be multiplied by to calculate work.

I hope this made sense and helped you out!
• At , wouldn't the time it takes for the box to fall the same because gravity is the only acceleration that makes the box fall. And the system is frictionless. • at
what if the object didn't get displaced because of friction would there be work? • How does the earth perform work (MGH)on one of your incline problems if the box was not moving in the direction of gravity? • Any path, inclined or curved, can be broken into small vertical and horizontal segments like the base and perpendicular of a right triangle with the short curved path segment substituting for the hypotenuse. As you move down the incline, you're moving both vertically and horizontally. The vertical displacements lead to a change in the gravitational potential energy while the horizontal displacements lead to zero change in U.
• at why did you say that the velocity slows down • If you recall that the energy in the system must remain constant, then we can say that the Kinetic Energy will always equal the Gravitation Potential Energy.

The the block goes up the ramp, the change in height is positive so GPE=mgh will increase.

The only way that GPE can increase is if Kinetic Energy Decreases.

The equation for Kinetic Energy is KE=1/2mV^2. So if (1/2) and (m) are both constants, the only way KE can decrease is if velocity decreases.

Another way of thinking about this is to visualize the system with a ball, the ball won't continue to maintain its speed as it goes up the hill as a result of gravity doing work on it.
(1 vote)
• At can you also use the formula work=force*displacement*cos theta? Would the times be the same if you were to do that?
(1 vote) • @Andrew M - I tried it:
P = W/T
P = F*d*cosθ / T
P = Fg*d*cosθ / t
P1 = (M*g)*H*cos2θ / T
P2 = (M*g)*H*cosθ / T

But in this case, unless I've done something wrong, the work on the second box seems to be greater as the cosine of a smaller angle is greater than that of a larger one. Even with the knowledge that the box on the steeper incline would reach the bottom in a shorter amount of time, I don't know how the power on that box would compare to the power on the second one due to the increased work on the second box.

If you could tell me where I messed up, it would be greatly appreciated.

EDIT: Did I perhaps forget to account for the fact that the net force on a box on an incline is across the parallel component?
• at Does that mean that, if the final velocity of that object were -4 m/s (still going in the same direction) as opposed to 4 m/s as was seen in the video, the net work would still be the same? That doesn't really make sense to me, but mathematically it works out for whatever reason.
i mean, doesn't it take a lot of effort(energy) to slow down the object and then accelerate it again to 4m/s on the opposite direction rather than slow down it a little bit so that it moves 4m/s to the left?   