question_answer

The percentage increase in kinetic energy when velocity of the body is doubled is ____.

Answer:

Given, \[{{v}^{2}}=2{{v}_{1}}\] Percentage change in K.E. = ? \[K.E{{.}_{1}}=?K.E{{.}_{2}}=?\] Change in \[K.E.=?\] We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] \[K.E{{.}_{1}}=\frac{1}{2}mv_{1}^{2}K.E{{.}_{2}}=\frac{1}{2}mv_{2}^{2}=\frac{1}{2}m{{(2{{v}_{1}})}^{2}}\]                                                 \[=4\left( \frac{1}{2}mv_{1}^{2} \right)\]              \[\therefore \]Change in \[K.E.=K.E{{.}_{2}}-K.E{{.}_{1}}=4\,\,K.E{{.}_{1}}-\] \[K.E{{.}_{1}}=3\,\,K.E{{.}_{1}}\] Percentage change in K.E. \[=\frac{change\,\,in\,\,K.E.}{initial\,\,K.E.}\times 100\] Therefore, percentage change in K.E is 300%.