A child bounces a 48 g superball on the sidewalk. The velocity change of the superball is from 28 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

F = 1.2×10⁻³ N

Explanation:

From the question,

Applying newton's second law of motion,

F = m(v-u)/t Equation 1

Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of  contact.

Note: let downward be negative and upward be positive.

Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = -28 m/s (downward),

t = 1800 s

Substitute into equation 1

F = 0.048(17-[28])/1800

F = 1.2×10⁻³ N

answer; equivalent to a one(1) horsepower motor;