Question

The speed of a moving body is halved. What change takes place in its kinetic energy?

Answer 1

Step 1: Assumption

Let the velocity of the body is '$\mathrm{v}$',

According to the question, the new velocity of the body is $\mathrm{v}\text{'}=\frac{\mathrm{v}}{2}$, and

To find, Change in kinetic energy

Step 2: Formula used

$\mathrm{K}.\mathrm{E}=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2$

Step 3: Calculation

Original Kinetic energy, $\mathrm{K}.\mathrm{E}=\frac{1}{2}\mathrm{x}\mathrm{m}\mathrm{x}{\mathrm{v}}^2$……… (1)

New Kinetic energy, $\mathrm{K}.\mathrm{E}\text{'}=\frac{1}{2}\mathrm{x}\mathrm{m}\mathrm{x}{(\mathrm{v}\text{'})}^2$……….(2)

Dividing equation (2) by (1)

$\frac{\mathrm{K}.\mathrm{E}\text{'}}{\mathrm{K}.\mathrm{E}}=\frac{\frac{1}{2}\mathrm{x}\mathrm{m}\mathrm{x}{(\mathrm{v}\text{'})}^2}{\frac{1}{2}\mathrm{x}\mathrm{m}\mathrm{x}{\left(\mathrm{v}\right)}^2}=\frac{\frac{1}{2}\mathrm{x}\mathrm{m}\mathrm{x}{(\mathrm{v}/2)}^2}{\frac{1}{2}\mathrm{x}\mathrm{m}\mathrm{x}{\left(\mathrm{v}\right)}^2}=\frac{1}{4}$

$\mathrm{K}.\mathrm{E}\text{'}=\frac{\mathrm{K}.\mathrm{E}}{4}$

Step 4: Final answer

Hence, the kinetic energy is decreased by four times.