Homework Help: Physics: skier problem

1. Oct 13, 2009

triplel777

1. The problem statement, all variables and given/known data

A 65-kg skier coasts up a snow-covered hill that makes an angle of 25° with the horizontal. The initial speed of the skier is 6.4 m/s. After coasting 1.9 m up the slope, the skier has a speed of is 3.2 m/s.

2. Relevant equations

3. The attempt at a solution

Work done if force F applied over a distance d.
W= Fd
Also
Ke= 0.5mV^2
Difference in Kinetic energy dKe
dKe=Ke1 - Ke2
dKe= 0.5m(V1^2 - V2^2)
dKe= W

a) W=dKe= 0.5m(V1^2 - V2^2)
W= 0.5x65(6.4^2 - 3.2^2)
W= 998.4 Joules

P=mgh
Pe= 65 x 9.81 x 1.9 sin(25)= 511.5 J

Wf= W- Pe= 998.4 - 511.5 = 486.9 J

what am i doing wrong?

2. Oct 13, 2009

tiny-tim

Hi triplel777!
I'm confused

what's the question?

3. Oct 13, 2009

triplel777

(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?

4. Oct 14, 2009

tiny-tim

Hi triplel77!

(try using the X2 tag just above the Reply box; and we usually write KE and PE with capitals, and joules without )
Apart from making PE = 512.0 J, I get the same work done by friction as you do.

5. Oct 14, 2009

triplel777

huh... ok thanks

6. Oct 15, 2009

triplel777

my teacher said to use this formula
Wnc= delta KE+ delta PE
would this make any difference?

7. Oct 15, 2009

gamer_x_

delta PE is just your change in gravitational potential energy, delta KE is your change in kinetic energy. Does this formula seem different than the one you used?