How Is Kinetic Friction Calculated in Skier's Uphill Motion?

[onyxorca]

Homework Statement

A 64.9-kg skier coasts up a snow-covered hill that makes an angle of 23.4 ° with the horizontal. The initial speed of the skier is 8.95 m/s. After coasting a distance of 2.31 m up the slope, the speed of the skier is 4.40 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?

http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c06/EAT_12258936060120_435296769488702.gif

Homework Equations

W=FX 2K=mv^2 a=(v^2-v0^2)/2x

The Attempt at a Solution

for (b) i used f=ma-mg sinθ,

or 1/2mv^2-1/2mv0^2-mgxsinθ for (a),

but neither was right...

 

[Delphi51]

1/2mv^2-1/2mv0^2-mgxsinθ for (a)
looks good to me. What did you get?

 

[dr_k]

1/2mv^2-1/2mv0^2-mgxsinθ for (a)
looks good to me. What did you get?

I think there may be a sign error here?

$\rm \Delta E = K_f - K_i + U_{f_{grav}}-U_{i_{grav}} = \frac{1}{2}m (v_f^2-v_i^2) +mg s Sin\theta = -1390J$

 

[onyxorca]

Correct! thank you!

 

[dr_k]

Correct! thank you!

You're welcome. If I had $1 for every time I've made a sign error in my lifetime, well, I'd own beach-front property in South Hampton.