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.
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?
1 answer
2 answers