The speed of a body is reduced to half of its original value. By what factor does this change the kinetic energy of the body?

Step 1: Given

Let the mass of the body $=m$

The original velocity of the body $v_o=v$

The half velocity of the body will be $v_h=\frac{v}{2}$

Step 2: Calculation of KE

KE at original velocity

$K{E}_o=\frac{1}{2}m{v_o}^2=\frac{1}{2}m{v}^2----\left(i\right)$

KE at half Velocity

$K{E}_h=\frac{1}{2}m{v}_h^2=\frac{1}{2}\times m\times (\frac{v}{2}{)}^2=\frac{1}{2}\times m\times \frac{v^2}{4}=\frac{K{E}_o}{4}----(ii)$

Hence, If the velocity is reduced to it half of the original value then kinetic energy will be reduced to 4 times.