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

`F=1.02x10^(-3) N`

Step-by-step explanation:

From the exercise we know:

`m=51g*(1kg)/(1000g)=0.051kg`

`v_(1)=-22m/s`

`v_(2)=14m/s`

`t_(2)-t_(1)=1800s`

So, the average acceleration is:

`a=(v_(2)-v_(1))/(t_(2)-t_(1))=((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`