Question

The coefficient of static and kinetic friction between a 3.0 kg box and a table are 0.40 and 0.30 respectively. What is the net force on the box when a 10.0 N horizontal force is applied to the box when at rest on a table?

Answer 1

The static friction $F_s=\mu_s mg=0.40\times 3.0\times 9.8=11.8\:\rm N.$

The kinetic friction $F_k=\mu_s mg=0.30\times 3.0\times 9.8=8.82\:\rm N.$

Since the applied force is less than static friction, so the box will be at rest, and the net force is zero.