Question

A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 24 m/s downward to 11 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? Answer in units of N.

Answer 1

Correct question:

A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 24 m/s downward to 11 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? Answer in units of N.

Answer:

The magnitude of the average force exerted on the superball by the sidewalk is 592.8 N

Step-by-step explanation:

Given;

mass of the superball, m = 57 g = 0.057 kg

initial velocity of the superball, u = 24 m/s

final velocity of the superball, v = 11 m/s

contact time with the sidewalk, t = 1 / 800 s

To determine the magnitude of the average force exerted on the superball by the sidewalk, we apply Newton's second law of motion;

F = ma

`But, a = (v-u)/(t) \\\\Thus, F = m((v-u)/(t) )\\\\F = 0.057((11-24)/(1/800) )\\\\F = 0.057((-13)/(1/800))\\\\F = -0.057((800*13)/(1))\\\\ F = -592.8 \ N\\\\Magnitude \ of \ the \ force \ is \ 592.8 \ N`

Therefore, the magnitude of the average force exerted on the superball by the sidewalk is 592.8 N