Question

If the velocity of a body is doubled, then what is the percentage increase in its kinetic energy?

need step by step explanation plzz...

Answer 1

Kinetic energy is doubled, that means the velocity increases by √2 since,

K.E= 1/2 m v^2

So the change in linear momentum will be

√2mv-mv

% change is (√2–1) * 100

thanks....

Answer 2

As we know that :

$k = \frac{1}{2} m {v}^{2}$

Where, k = kinetic energy,

m = Mass of the body,

and v = Velocity of the body

Then, the kinetic energy of the body whom mass is 'm' and velocity is 'v' will be :

$k_{1} = \frac{1}{2} m {v}^{2}$

When the velocity will be doubled then, its kinetic energy will be :

$k_{2} = \frac{1}{2} m {(2v)}^{2} \\ \\ = > k_{2} = \frac{1}{2} m4 {v}^{2} \\ \\ = > k_{2} = 4. \frac{1}{2} m {v}^{2} \\ \\ = > k_{2} = 4 k_{1}$

Now, percentage change :

$\frac{4 k_{1} - k_{1} }{ k_{1}} \times 100 \\ \\ = > \frac{3 k_{1} }{ k_{1} } \times 100 \\ \\ = > 300$

So, the new kinetic energy will be 300% increased.