Question

The velocity of a body is reduced to half of its original value. Find the percentage decrease in its K.E.

Answer 1

Decrease in kinetic energy, $\Delta K.E=\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(\frac{v}{2}\right)}^2=\frac{3}{4}\left(\frac{1}{2}m{v}^2\right)$

Percentage decrease is $\frac{\Delta K.E}{K.E}\times 100=\frac{\frac{3}{4}\left(\frac{1}{2}m{v}^2\right)}{\frac{1}{2}m{v}^2}\times 100=\frac{3}{4}\times 100=75percent$

Hence, $75$ % kinetic energy will decrease.