Question

The shoe is in contact with the initially nearly stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football in Newtons to give it a velocity of 22 m/s?

Answer 1

If v=22m/s after0.02s then acceleration was
22/0.02=1100m/s^2
Now, F=ma=(0.5)(1100)=550N

Answer 2

F = 550 N

Step-by-step explanation:

As we know that acceleration of the football is given as

`a = (v_f - v_i)/(\Delta t)`

here we know that

`v_i = 0`

`v_f = 22 m/s`

`\Delta t = 20 ms`

now we have

`a = (22 - 0)/(20 * 10^(-3))`

`a = 1100 m/s^2`

now the force exerted on the ball is given as

`F = ma`

`F = 0.5 * 1100`

`F = 550 N`