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A pool cue striking a stationary billiard ball (mass=0.25kg) gives the ball a speed of 2 m/s. If the force of the cue on the ball was 25N, over what distance did this force act? |
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Answer» Solution :The kinetic energy of the ball as it LEAVES the cue is `K=(1)/(2)MV^(2)=(1)/(2)(0.25kg)(2m//s)^(2)=0.5J` The WORK W done by the cue gave the ball this kinetic energy, so `W=DeltaKimpliesW=K_(i) implies Fd=Kimplies d=(K)/(F)=(0.5J)/(25N)=0.02m=2cm` Note that this could have been solved by using `F_("net")=ma` to find the acceleration, and them using Big Five +5 to find the displacement. |
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