Question

A child bounces a 51 g superball on the sidewalk. The velocity change of the super bowl is from 22 m/s downward to 14 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 1

Answer:

$F=1.02x10^{-3} N$

Explanation:

From the exercise we know:

$m=51g*\frac{1kg}{1000g}=0.051kg$

$v_{1}=-22m/s$

$v_{2}=14m/s$

$t_{2}-t_{1}=1800s$

So, the average acceleration is:

$a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{(14-(-22))m/s}{1800s}=0.02m/s^2$

The average force is:

$F=m*a=(0.051kg)(0.02m/s^2)=1.02x10^{-3} N$