Question #22833

A 0.43 kg football is thrown with a velocity
of 12 m/s to the right. A stationary receiver
catches the ball and brings it to rest in 0.023
s.
What is the force exerted on the receiver?
Answer in units of N

Expert's answer

F=m*a, second newton's law.

F=0.43*a -& exerted force

a=v/t=12*0.023 - we find the acceleration, that stopped the ball while force was exerted

So F=0.43*12/0.023=224 N - we put in numbers to find the answer

F=0.43*a -& exerted force

a=v/t=12*0.023 - we find the acceleration, that stopped the ball while force was exerted

So F=0.43*12/0.023=224 N - we put in numbers to find the answer

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