Answer to Question #22833 in Mechanics | Relativity for matt

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
1
Expert's answer
2013-01-24T09:52:53-0500
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

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