Question

A factory worker exerts a 100 N horizontal pushing force on a 70.0 kg box that is initially at rest on the floor. The coefficient of static friction is 0.2 and the coefficient of kinetic friction is 0.1. The force of friction acting on the box is equal to:

Answer 1

As block is initially at rest. So , in order to move it, applied force should be greater than limiting static friction force. Limiting value of static friction force here is given by

= 0.2×70×9.8

= 137.2 N

Since applied force is less, box will remain at rest. Static friction force of 100 N will act to balance the applied force (so that net force in horizontal direction is zero) and to keep the box at rest.

f = 100 N, in direction opposite to applied force.