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
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
[tex]But, a = \frac{v-u}{t} \\\\Thus, F = m(\frac{v-u}{t} )\\\\F = 0.057(\frac{11-24}{1/800} )\\\\F = 0.057(\frac{-13}{1/800})\\\\F = -0.057(\frac{800*13}{1})\\\\ F = -592.8 \ N\\\\Magnitude \ of \ the \ force \ is \ 592.8 \ N[/tex]
Therefore, the magnitude of the average force exerted on the superball by the sidewalk is 592.8 N
