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A 7-kg bowling ball moving at 4 m/s strikes a 1 kg bowling pin. If the ball slows to 2 m/s in 0.05 s, how much force does it exert on the pin?

Please, show your work.

Respuesta :

Answer:

280 N

Explanation:

acceleration = v2-v1 / time taken = (2-4 )/ 0.05 = -40 m/s^2  ( neg sign indicates slowing down )

force exerted = ma = 7 kg x -40 m/s^2 = - 280 N ( neg sign means opposite direction of initial velocity )

since the 7 kg ball is slowing down, the  direction of force will be opposite of the initial velocity , and it will be 280 N

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