Question

If a man increases his speed by $2m/s$, his kinetic energy is doubled. The original speed of the man is

• A. $(2+\sqrt{2})m/s$
• B. $(2+2\sqrt{2})m/s$
• C. $4m/s$
• D. $(1+\sqrt{2})m/s$

Answer 1

The correct option is B $(2+2\sqrt{2})m/s$

Given that velocity is increased from $v$ to $v+2$.

Kinetic energy
$K{E}_i=\frac{1}{2}m{v}^2$
$K{E}_f=\frac{1}{2}m(v+2{)}^2$

Given
$\Rightarrow K{E}_f=2K{E}_i$
$\Rightarrow \frac{1}{2}m(v+2{)}^2=2\times \frac{1}{2}m{v}^2$
$\Rightarrow v+2=\sqrt{2}v$
$\Rightarrow v(\sqrt{2}-1)=2$
$\Rightarrow v=\frac{2}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$
$\Rightarrow v=(2\sqrt{2}+2)m/s$