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  4. When a man increases his speed by 2 ms -1, he finds that his kinetic energy is doubled, the original speed of the man is

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Q. When a man increases his speed by $2 \,ms ^{-1}$, he finds that his kinetic energy is doubled, the original speed of the man is

Work, Energy and Power

Solution:

$KE =\frac{1}{2} m v^{2}$
Given, $v_{2}=\left(v_{1}+2\right)$
$ \therefore \frac{K_{1}}{K_{2}} =\left(\frac{v_{1}}{v_{2}}\right)^{2} $
$\Rightarrow \frac{1}{2} =\frac{v_{1}^{2}}{\left(v_{1}+2\right)^{2}} $
$\left(\therefore K_{2}=2 K_{1}\right)$
$v_{1}^{2}+4 v_{1}+4 =2 v_{1}^{2} $
$ v_{1}^{2}-4 v_{1}-4 =0 $
$v_{1}=\frac{4 \pm \sqrt{16+16}}{2}$
$v_{1}=\frac{4+\sqrt{32}}{2}$
$=2(\sqrt{2}+1) \,ms ^{-1}$