Question #273553

A basketball is shot through the air, hitting the backboard of a hoop. During the shot, the basketball was travelling at 8 m/s, and right after hitting the backboard, the basketball travels 1 m/s in the opposite direction. The mass of a standard basketball is 590g.



a. Find the kinetic energy of the basketball before hitting the backboard (2pts)


b. Find the kinetic energy of the basketball after hitting the backboard (2pts)


c. Find the work done by the backboard to the basketball (2pts)


Expert's answer

(a)

KEi=12mvi2,KE_i=\dfrac{1}{2}mv_i^2,KEi=12×0.59 kg×(8 ms)2=18.88 J.KE_i=\dfrac{1}{2}\times0.59\ kg\times(8\ \dfrac{m}{s})^2=18.88\ J.

(b)

KEf=12mvf2,KE_f=\dfrac{1}{2}mv_f^2,KEf=12×0.59 kg×(1 ms)2=0.295 J.KE_f=\dfrac{1}{2}\times0.59\ kg\times(1\ \dfrac{m}{s})^2=0.295\ J.

(c)

W=ΔKE=KEfKEi,W=\Delta KE=KE_f-KE_i,W=0.295 J18.88 J=18.585 J.W=0.295\ J-18.88\ J=-18.585\ J.

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