# A man increases his speed by 2m/s, then his kinetic energy gets doubled. The original speed of the man is :

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A man increases his speed by 2m/s, then his kinetic energy gets doubled. The original speed of the man is :
1. √2(√2 + 1) m/s
2. √2(√2 - 10)m/s
3. 2(√2 + 1)m\s
4. 2(√2 - 1) m/s

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Correct Answer - Option 3 : 2(√2 + 1)m\s

CONCEPT:

Kinetic energy

• The energy possessed by a body due to the virtue of its motion is called kinetic energy.

$\Rightarrow KE=\frac{1}{2}mv^{2}$

Where KE = kinetic energy, m = mass and v = velocity

CALCULATION:

Given Δv = v2 - v1 = 2m/s and KE2 = 2KE1

Where Δv = increase in speed

• Initial kinetic energy,

$\Rightarrow KE_{1}=\frac{1}{2}mv_{1}^{2}$     -----(1)

• Final kinetic energy,

$\Rightarrow KE_{2}=\frac{1}{2}mv_{2}^{2}$     -----(2)

By equation 1 and equation 2,

$\Rightarrow KE_{2}=2KE_{1}$

$\Rightarrow \frac{1}{2}mv_{2}^{2}=2\times\frac{1}{2}mv_{1}^{2}$

$\Rightarrow v_{2}^{2}=2v_{1}^{2}$

$\Rightarrow v_{2}=\pm\sqrt{2}v_{1}$     -----(3)

Since,

$\Rightarrow \Delta v = v_{2}-v_{1}=2m/s$

By equation 3,

$\Rightarrow \sqrt{2}v_{1}-v_{1}=2m/s$

$\Rightarrow v_{1}=\frac{2}{\sqrt{2}-1}$     -----(4)

By dividing and multiplying $\sqrt{2}+1$ in equation 4,

$\Rightarrow v_{1}=\frac{2}{\sqrt{2}-1}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}$

$\Rightarrow v_{1}=\frac{2(\sqrt{2}+1)}{\sqrt{2}^2-1^2}$

$\Rightarrow v_{1}=2(\sqrt{2}+1)$m/s

• Hence, option 3 is correct.