How Do You Calculate the Work Done by Friction on a Skier?

[neoncrazy101]

Homework Statement

A 71.2-kg skier coasts up a snow-covered hill that makes an angle of 31.4 ° with the horizontal. The initial speed of the skier is 9.96 m/s. After coasting a distance of 2.04 m up the slope, the speed of the skier is 4.41 m/s. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Homework Equations

Ef= Eo (aka 1/2mv_f^2 + mgh_f = 1/2mv_o^2 + mgh_o

The Attempt at a Solution

Final: 1/2(71.2)(4.41)²+71.2(-9.8)(1.74) = -521 (H=2.04cos(31.4), right?)
Initial: 1/2(71.2)(9.96)²+71(-9.8)(0) = 3531

The answer our phyrics program gives for this problem is A: -2100J and for B: 1030N. But I cannot figure this problem out. What am I doing wrong?

 

[jhae2.718]

The angle is with the horizontal, so the height should be 2.04sin(31.4)

 

[neoncrazy101]

Ok, that was one mistake but I must of done something else, I still get the wrong answer.

Final: 1/2(71.2)(4.41)²+(71.2)(-9.8)(1.06) = -47.27324
Initial: 1/2(71.2)(9.96)²+(71.2)(-9.8)(0) = 3531.57696

Ef - Eo = -3578.85284

The answer is supposed to be -2100J. Must be something else that I am doing wrong.

 

[HallsofIvy]

Your sign on g is wrong. The way you have it, the potential energy is lower when you are higher up.

 

[neoncrazy101]

Oh. that's works. I totally thought that it was supposed to be -9.8. thanks. Now I'm getting -2100.

 

[jhae2.718]

For energy, g is almost always positive. In fact, it's best to leave g positive and determine the sign based on the coordinate system in use.