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

$$W=W_f+K\\Fs\cos{20}=\mu Ns+K\\\\K=Fs\cos{20}-\mu (mg-F\sin{20})s\\K=65.5(16.2)\cos{20}\\-0.19 ((3.333)(9.8)-65.5\sin{20})(16.2)=966\ J$$