Question #78474
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?
Expert's answer
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
(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
