#### Question

A box with a weight of mg = 50 N is at rest on a flat surface. A force of 25 N directed right is then applied. If the coefficient of static friction between the box and the surface is 0.6, and the coefficient of kinetic friction between the box and the surface is 0.4, what is the force of friction acting on the box?

- 20 N to the left
- 25 N to the left
- 30 N to the left

The applied force is not enough to overcome the force of static friction, so the box remains at rest. To give a net force of zero, the force of friction exactly balances the applied force, so it is 25 N to the left.

In the simulation a force directed right is applied. Its magnitude is steadily increased from zero until the box starts to move, and then the force is maintained at that level. The frictional force balances the applied force exactly until the maximum static friction force is reached. Beyond that the box starts to move so the friction force is the kinetic force of friction, which is less than the maximum static force of friction. Now there is a constant net force, so the box accelerates to the right.