Asked by Raf

A skier (75 kg) goes down a 15* slope. The coefficient of kinetic friction between the skis and the snow is 0.185. The length of the slope is of 500m.

a) What is the work done by gravity?

Answer) Wmg = mgd

= 75kg(9.8N/kg)500m(sin15)
= 354 977 J

Is it cos or sin? And is Work done by both air and friction supposed to be higher than work done by gravity at the end of the slope?
Answered by drwls
The sine is correct function to use, but you made a mistake in the multiplication.

Work done by friction (on snow) and aerodynamic drag is negative. The skier does the work. The absolute value equals the gravity work only if the kinetic energy does not change.
Answered by Raf
what is the mistake? And thank you for the information!
Answered by Raf
and drag and friction work has to be higher at the end than gravity in order to stop the skier at the bottom, right?
Answered by Raf
oh and I see what you mean, it's 95 116 J, thanks again!
Answered by drwls
75*9.8*500*(sin15) = 9.516*10^4 J
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