Question

A force of 65.5 N is applied at an angle of 20.0 (measured from the horizontal) to slide a 3.33 kg block 16.2m across the floor. If the coefficient of friction is 0.190, calculate the final kinetic energy.

Answer 1

$$N=F\sin{20}-mg\\ K=W=(F\sin{20}-\mu N)s\\K=(F\sin{20}-\mu (F\sin{20}-mg))\\K=(65.5\sin{20}-0.19 (65.5\sin{20}-(3.33)(9.8)))\\K=24.3\ J$$