Question

The coefficients of static and kinetic friction between a 3.0-kg box and a horizontal desktop are 0.40 and 0.30, respectively. What is the force of friction on the box when a 15-N horizontal push is applied to the box?

Answer 1

The force of friction on the box is 8.82 N.

Explanation:

The available static friction is given by

`F_"sf" = mu_s*N = mu_s*m*g`

`F_"sf" = 0.4*3.0 kg*9.8 m/s^2 = 11.8 N`

11.8 N of static friction is not enough to prevent 15 N from moving this box. So we need to determine the kinetic friction.

`F_"kf" = mu_k*N = mu_k*m*g`

`F_"kf" = 0.3*3.0 kg*9.8 m/s^2 = 8.82 N`

That is the "force of friction on the box" that the question asked for.

(For information, the net force acting on the box is
`F_"net" = 15 N - 8.82 N = 6.18 N`
Any calculation of acceleration should use `F_"net"`.)

I hope this helps,
Steve