Question

9. A force is applied horizontally to a 20 kg box on a flat table, if the acceleration of the box is 2 m/s and the

coefficient of friction between the box and the table is 0.5, what was the magnitude (number) of the
applied force?

Answer 1

By Newton's second law,

n + (-w) = 0

p + (-f ) = (20 kg) (2 m/s²)

where n is the magnitude of the normal force, w is the weight of the box, p is the magnitude of the applied force (p for push or pull), and f is the magnitude of the friction force.

Calculate the weight of the box:

w = (20 kg) (9.80 m/s²) = 196 N

Then

n = w = 196 N

and

f = µ n = 0.5 (196 N) = 98 N

Now solve for p :

p - 98 N = 40 N

p = 138 N