How will the kinetic energy of an object change if its speed is halved?

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Solution

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$K i n e t i c e n e r g y o f t h e b o d y i s w h e n i t s s p e e d i s v K_{1} = \frac{1}{2} m v^{2} K i n e t i c e n e r g y o f t h e b o d y i s w h e n i t s s p e e d i s \frac{v}{2} K_{2} = \frac{1}{2} m (\frac{v}{2})^{2} = \frac{m v^{2}}{8} \frac{K_{2}}{K_{1}} = \frac{\frac{m v^{2}}{8}}{\frac{1}{2} m v^{2}} = \frac{1}{4} K_{2} = \frac{K_{1}}{4} R e g a r d s$

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