A man increases his speed by 2 m/s, and as a | vedclass

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(C) Let the initial speed be $u$ and the mass of the man be $m$.The initial kinetic energy is $E_1 = \frac{1}{2}mu^2$.The final speed is $(u + 2) \ m/

Vedclass · Accepted Answer

$(2 + 2\sqrt{2}) \ m/s$

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(C) Let the initial speed be $u$ and the mass of the man be $m$.The initial kinetic energy is $E_1 = \frac{1}{2}mu^2$.The final speed is $(u + 2) \ m/s$.The final kinetic energy is $E_2 = \frac{1}{2}m(u + 2)^2$.Given that the kinetic energy doubles, $E_2 = 2E_1$.Substituting the expressions: $\frac{1}{2}m(u + 2)^2 = 2 	imes \frac{1}{2}mu^2$.Simplifying: $(u + 2)^2 = 2u^2$.Taking the square root on both sides: $u + 2 = \sqrt{2}u$.Rearranging for $u$: $\sqrt{2}u - u = 2 \implies u(\sqrt{2} - 1) = 2$.Solving for $u$: $u = \frac{2}{\sqrt{2} - 1}$.Rationalizing the denominator: $u = \frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2(\sqrt{2} + 1)}{2 - 1} = 2\sqrt{2} + 2 \ m/s$.

$A$ man increases his speed by $2 \ m/s$, and as a result, his kinetic energy doubles. What was his initial speed?

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