How Much Work Is Done Against Friction?

[DKPeridot20]

I've been working on this particular question for some time and am sure that it's very simple but I continue to get the wrong answer...

An 18.0 N weight slides down a rough inclined plane which makes and angle of 29.0 degrees with the horizontal. The weight starts from rest and gains a speed of 15.0 m/s after sliding 150 m. How much work is done against friction?

I am under the impression that
1) I need to find Wtotal by saying it equals 1/2mv^2 (note: because the weight is 18.0 N I devided it by gravity=9.81m/s to get kg) (I've done it both ways though)
2) Find Wgravity by saying it equals mgh
3) Use Wtotal = Wgravity + Wfriction rearranged to say Wfriction = Wtotal - Wgravity
4) vwala - the wrong answer (I know this has to be something easy, I'm just not getting it. Friction and I are not friends...)

Choices are:
a. 1100 J
b. 1500 J
c. -1500 J
d. 229 J

How should I be going about this?

 

[mezarashi]

Your equation with work there is a bit odd. Let's look at it another way.

Usually friction is pretty formidable for beginning students because it is hard to analyze its direction during complex motion. This question however, spares you from such pain. The way I would recommend to approach the problem is consider:

1. How fast would the mass be going without any friction? You know the inclination, you know the distance traveled, mgh = KE.

2. How fast is it actually going? 15m/s... this is lower than the value you calculated in part 1 isn't it?

3. Calculate the difference in your theoretical KE and the actual (given) KE.

4. So where did this kinetic energy go? Who stole it? Who's the culprit. I think you can handle from here ;)