# Potential Energy Help Needed!

1. Dec 17, 2005

### avb203796

How would you find the gravitational potenetial energy change for a box being slide down a slope? The weight of the box is 490N, the kinetic coefficient of friction is 0.12N, the slant of the slope is 45 degrees. I knwo how to find the forces acting ont he boxes and have calculated all of them but I have no clue how to solve for the potential energy. Please help!

2. Dec 17, 2005

### mukundpa

change in gravitational potential energy of a body of mass m is mgh, where h is the vertical height between the two positions.

3. Dec 17, 2005

### avb203796

but given just an angle how would I know the height?

4. Dec 17, 2005

### andrewchang

are you sure that's all the information you've been given?

5. Dec 17, 2005

### mukundpa

1. The vertical height should be given.
2. They may give the slant height and with that and the angle known we can calculate the vertical height.

Just check again the question otherwise it is not possible to caclculat the change in KE with given data.

6. Dec 17, 2005

### avb203796

THis was thwe actual complete question but I am just trying to solve the last part:

A 50.0 kg crate slides down a slope of 45 degrees for a distance of 200.0m. how much work will be done by gravity? How much work will be done by friction, if the kinetic coefficent of friction is 0.12? What speed will it attain at this point if it was originally at rest? By how much will its gravitaional potential energy change during the slide?

Again I have solve all of the problem except for the last question.

7. Dec 17, 2005

### mukundpa

here
the distance covered is given 200m
hence the vertical height fallen is 200 sin 45deg. = 2000*707 = 141.4 m.
now you can calculate mgh
remember the height is lost means the potential energy decreases(may be the answer is having a negative sign)

8. Dec 17, 2005

### vaishakh

How did you solve the question1?

9. Dec 17, 2005

### mukundpa

the distance covered down the slope is given = 200m hence the vertical component of this distance is 200 sin45deg. = 200*0.707
mgh is the loss in potential energy.(_ve)

10. Dec 17, 2005

### avb203796

I solved for the weight which was 490N, then I multiplied the weight times the 200m distance times cos(45degrees). y?

11. Dec 17, 2005

### avb203796

Ok thank you mukundpa!

12. Dec 17, 2005

### vaishakh

Just think o the coincidence of the first question with the last one. I think you should know the relation between energy and work. What is the definition of energy? So you anyway solved it before posting?

13. Dec 17, 2005

### mukundpa

ya you may solve it in that way too if you have calculated the speed at the end
loss in potential energy = gain in KE + work done against friction