- #1
The Merf
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ok guys, i got one that wants me to figure out how much work is done:
A skier of a mass 79.1 kg, starting from rest, slides down a slope at an angle of 38 degrees with the horizontal. The coefficient of kinetic friction, u, is 0.09. What is the net work (the net gain in kinetic energy) done on the skier in the first 8.9 s of decent?
now i don't want the answer, i just need help know what equations i need when.
I know it gives me m=79.1kg theta=38degrees coefficient of kinetic friction=.09 (not sure of unit) and t=8.9
how do i get how much work?
I have figured the mass in the verticle direction to be 48.69 (is this right?) by taking sin38=x/79.1 or 79.1sin38=x
A skier of a mass 79.1 kg, starting from rest, slides down a slope at an angle of 38 degrees with the horizontal. The coefficient of kinetic friction, u, is 0.09. What is the net work (the net gain in kinetic energy) done on the skier in the first 8.9 s of decent?
now i don't want the answer, i just need help know what equations i need when.
I know it gives me m=79.1kg theta=38degrees coefficient of kinetic friction=.09 (not sure of unit) and t=8.9
how do i get how much work?
I have figured the mass in the verticle direction to be 48.69 (is this right?) by taking sin38=x/79.1 or 79.1sin38=x
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