Posted by Kim on .
A 69.4-kg skier coasts up a snow-covered hill that makes an angle of 24.3° with the horizontal. The initial speed of the skier is 7.04 m/s. After coasting 1.78 m up the slope, the skier has a speed of 3.04 m/s. Calculate the work done by the kinetic frictional force that acts on the skis.
I've done this question 3 different times, CAPA doesn't accept my answer.
I've used Energy lost as friction = KElost - PEgained, getting 900 J. But this isn't right. I don't know what else to do.
PE gained= mg(sin24.3)d
friction work= unknown
Initial KE-finalPE=workdoneby friction+finalKE
1/2 m*7.04^2-m*g*1.78*sin24.3-1/2 m 3.04^2= work done be friction
1/2 *69.4*(7.04^2-3.04^2)-mg*1.78*sin24.3=workdone by friction.