Question

A 3.0 kg box is at rest on a table. The static friction coefficient us between the box and table is 0.40, and

the kinetic friction coefficient per is 0.10. Then, a 10 N horizontal force is applied to the box.

Answer 1

A 3.0 kg box is at rest on a table. The static friction coefficient μs​ between the box and table is 0.40, and the kinetic friction coefficient μk is 0.10. Then, 20 N horizontal force is applied to the box.

What is the best estimate of the magnitude of the box’s acceleration?

Answer: 5.7 m/s²

Answer 2

While the box is at rest, the net vertical force acting on it is

∑ F = n - w = 0

where n = magnitude of the normal force (table pushing up on the box) and w = weight of the box. So we have

n = w = mg = (3.0 kg) (9.8 m/s²) = 29.4 N

The coefficient of static friction between the table and box is 0.40, so the maximum magnitude of static friction is

f = 0.40 n = 0.40 (29.4 N) = 11.76 N

which is to say, the box will not move unless a force larger than this is applied to the box.

10 N is of course smaller than 11.76 N, so the box would stay at rest, and its acceleration would remain 0 m/s².