Question

A stationary 25 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.50 and of kinetic friction is 0.30. A horizontal force of 300 N is applied to the object. Determine the acceleration of the object. *

Answer 1

The net vertical force on the object is

∑ F[vertical] = n - mg = 0

where n is the magnitude of the normal force exerted by the surface, m is the object's mass, and g is the mag. of acceleration due to gravity. It follows that

n = mg = (25 kg) (9.8 m/s²) = 245 N

The net horizontal force is

∑ F[horizontal] = 300 N - f = ma

where f is the mag. of friction and a is the object's acceleration.

We have

f = µn

where µ is the coefficient of friction. Since the object starts at rest, it won't move and accelerate unless the applied force of 300 N is sufficient to overcome the maximum static friction, which is

f = 0.50 n = 0.50 (245 N) = 122.5 N

Since f < 300 N, the box will begin to slide, at which point the coefficient of kinetic friction kicks in and the mag. of friction is

f = 0.30 n = 0.30 (245 N) = 73.5 N

Now solve for a :

300 N - 73.5 N = (25 kg) a

a = (226.5 N) / (25 kg)

a = 9.06 m/s² ≈ 9.1 m/s²