Question

A 5.0 kg box slides up a 10 m long frictionless incline at an angle of 20 degrees with the horizontal, pushed by a 40 N force parallel to the incline. What is the change in kinetic energy?

Answer 1

I tried using 1/2mvi^2 = 1/2mvf^2 + mgh, but I'm not sure how to get vf and vi.

Answer 2

Unless we assume vi is 0?

Answer 3

@theEric

Answer 4

Hi! The change in kinetic energy is the work done!

Work is done by two forces, the pushing the block and the gravity on the block.

The change in kinetic energy is the work done.. I think the work done by the net force :)

Answer 5

Work is \(\vec F\bullet \vec d=F\ d\ \cos\theta\) where \(\theta\) is the angle between the force and displacement.

Answer 6

Good luck!

Answer 7

F*d = 40(10)(cos(20)), correct? That gives an incorrect answer...

Answer 8

Hi! I have to go soon, but I'll tell you this!

You have two phenomena doing work:
1. Your force
2. Gravitational force

1. To calculate your force, the angle \(\theta\) is \(0\) since it the force and displacement are parallel!
2. When you consider gravitational work, the easiest thing to do would be to consider that is the kinetic energy change. I mean, KE is decreased by the gravity as it pulls it down. But this is an increase in PE. The opposite of the PE is the KE. KE=-PE, because KE increases just as much as PE increases, so KE=-mgh. Or you could use the angle for that.

So you look at the work done from both.
1. Pushing the block gives KE by its work.
2. Gravity reduces the KE by its work.

If I'm right, the angle isn't important...

Answer 9

Let me know if I said something that doesn't make sense. I'll be studying, but I'll look up at my monitor every once in a while.

Answer 10

But I'll log out in a half hour.

Answer 11

Okay, thanks.

Answer 12

And I like at the beginning how you started the problem. Unfortunately we don't know anything about the velocities, or KE. But we can use what I talked about to find the CHANGE in KE.

It's called the Work-Kinetic Energy theorem, commonly.

Answer 13

\[x^2\]