Your \"friend\" throws a 5.4 kg bowling ball at you. Let\'s think about how you

Question

Your "friend" throws a 5.4 kg bowling ball at you. Let's think about how you would catch it to prevent breaking a bone? You would stick your arms out toward the bowling ball, and bring your arms closer to you as you start to catch it, to maximize the distance you are slowing it down. Assume the ball is traveling at 3.9 m/s just as it reaches your hands, and your force is the only thing acting on it in the direction of motion (so all done in the horizontal direction). If your arms slow the ball over a distance of 0.27 m, what is the magnitude of the average force you exert on the bowling ball?

Explanation / Answer

let

d = 0.27 m/s

m = 5.4 kg

d= 0.27 m

let F is force applied by the hand on ball.

Apply Work-energy throem

Workdone by you on ball = change in kinetic energy of the ball

F*d*cos(180) = KEf - KEi

-F*d = -KEi (since KEf = 0)

F = KEi/d

= 0.5*m*vi^2/d

= 0.5*5.4*3.9^2/0.27

= 152.1 N

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