Answer to Question #117483 in Physics for Higgins

Question

A clerk pushes a cabinet of mass 22.0 kg across the floor by exerting a horizontal force of magnitude 98 N. The
coefficient of friction is 0.2 between the floor and the cabinet. If the cabinet starts from rest, determine the
speed of the cabinet after it has moved 1.2 m.

Answer 1

$$a=\frac{F-\mu mg}{m}=\frac{98-(0.2)(22)(9.8)}{22}=2.5\frac{m}{s^2}$$

$$v^2=2as\to v^2=2(2.5)(1.2)$$

$$v=2.4\frac{m}{s}$$