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A 2.0-kg softball is pitched to you at 20 m/s. You hit the ball back along the same path, and at the same speed. If the bat was in contact with the ball for 0.10 s, what is the magnitude of the average force the bat exerted?

Respuesta :

Answer: 800 N

Explanation:

Force is change of momentum (p) over change in time: F = ma = dp/dt

change in time dt= 0.10 s

mass = 2 kg

initial velocity v0=20 m/s

final velocity vf= -20 m/s

change in momentum dp= m(vf-v0)

dp=2(-20-20)= -80 N/s

F=dp/dt=-80/.10=-800 N

Since they only want the magnitude, you can drop the negative sign so the answer is just 800 N.

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