# Conservation of Energy crate and ramp

1. Oct 18, 2007

### Dynex

A factory worker accidentally releases a 230 kg crate that was being held at rest at the top of a ramp that is 4.3 m long and inclined at 40° to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is 0.37. (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate's kinetic energy does not change as it moves from the ramp onto the floor.)

Thankz for the help in advance!!

2. Oct 18, 2007

### Yannick

Hi Dynex

How did you try to solve it?

My approach would be:

1. Draw the situation!
2. Calculate the velocity of the mass at the end of the ramp
3. With the given friction and the velocity from 2. you should be able to get the distance the crate slides on the floor.

Yannick

P.S. I hope my English is understandable...

Last edited: Oct 18, 2007
3. Oct 18, 2007

### Bill Foster

In order to find v, you'll need to find a. Then you can apply this equation to find v:

a(x-x0)=½(v²-v0²)

x0 and v0 will be 0. So...

ax=½v²

To find a, add up all the force vectors acting on the box (treated as a point).

F=ma=Wsin(θ)-μN where W=mg and N=Wcos(θ)=mgcos(θ)
=mgsin(θ)-μmgcos(θ)
=mg(sin(θ)-μcos(θ))

so...

a=g(sin(θ)-μcos(θ))