Answer to Question #78474 in Physics for Bella June Dalal

Question

A 0.107 kg puck is initially moving at 21.2m/s[W]along the ice. The coefficient of kinetic friction for the puck and the ice is 0.005. What is the speed of the puck after travelling 58.5m?

Answer 1

The change of kinetic energy is equal to friction force work done
(mv_f^2)/2-(mv_i^2)/2=-F_frict l=-fmgl
Thus
v_f=√(v_i^2-2fgl)
=√(〖21.2〗^2-2×0.005×9.81×58.5)
=21.1 m/s