Question

A constant unbalanced force of friction acts on a 15.0-kilogram mass moving along a horizontal surface at 10.0 meters per second. If the mass is brought to rest in 1.50 seconds, what is the magnitude of the force of friction

Answer 1

The mass slows to a rest from an initial speed of 10.0 m/s in a matter of 1.50 s. If we assume constant acceleration, then the mass has acceleration a such that

0 = 10.0 m/s + a (1.50 s)   ⇒   a ≈ -(10.0 m/s)/(1.50 s)

The net force acting on the mass as it slows down is

∑ F[horizontal] = -F[friction] = ma

where m = 15.0 kg, and we take the direction in which friction is acting to be negative. Then

-F[friction] = -(15.0 kg) (10.0 m/s)/(1.50 s)   ⇒   F[friction] = 100 N