Question

A 50.0N box is at rest on a horizontal surface. The coefficient of static friction between the box and the surface is 0.50, and the coefficient of kinetic friction is 0.30. A horizontal 20.0N force is then exerted on the box. Find the magnitude of the acceleration of the box

Answer 1

0 m/s²

Step-by-step explanation:

Draw a free body diagram of the box. There are four forces on the box:

Weight force mg pulling down.

Normal force Fn pushing up.

Applied force F pushing right.

Friction force Fn μ pushing left.

Sum the forces in the y direction.

∑F = ma

Fn − mg = 0

Fn = mg

Find the static friction:

Fs = Fn μs

Fs = mg μs

Fs = (50.0 N) (0.50)

Fs = 25.0 N

The applied force of 20.0 N is not enough to overcome the static friction. So the box does not move.