Asked by Will
A puck of mass 0.165 kg, initially resting on ice, is hit by a stick to gain a speed of 32.0 m/s and then slides 25.0 m to lose 5.00% of its kinetic energy. (a) What is the work done on the puck by the stick? (b) What is the kinetic friction coefficient between the ice and puck?
Answers
Answered by
bobpursley
work done on puck=1/2 masspuck*v^2
initiial KE-finalKE=mg*mu
1/2 m vi^2-1/2 (.95vi)^2=mg*mu
solve for mu.
initiial KE-finalKE=mg*mu
1/2 m vi^2-1/2 (.95vi)^2=mg*mu
solve for mu.
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