Question

A ball dropped from a height of $4\mathrm{m}$ rebounds to a height of $2.4\mathrm{m}$ after hitting the ground. Then the percentage of energy lost is:

• A. $40$
• B. $50$
• C. $30$
• D. $600$

Answer 1

The correct option is A

$40$

Step 1: Given data

If an object has mass $\text{'}\mathrm{m}\text{'}$ and is lifted through a height $\text{'}\mathrm{h}\text{'}$from the surface of earth, then the potential energy of an object will be:

${\mathrm{E}}_{\mathrm{p}}=\mathrm{mgh}$

Here, $\text{'}\mathrm{g}\text{'}$ is acceleration due to gravity.

Let us consider the height of a ball before hitting the ground, ${\mathrm{h}}_1=4\mathrm{m}$

Height of a ball after hitting the ground, ${\mathrm{h}}_2=2.4\mathrm{m}$

Step 2: Calculating the percentage of energy lost

Let potential energy of a ball before hitting the ground, ${\mathrm{E}}_1={\mathrm{mgh}}_1$

Potential energy of a ball after hitting a ground and rebounding to height $\text{'}{\mathrm{h}}_2\text{'}$ is, ${\mathrm{E}}_2={\mathrm{mgh}}_2$

Percentage of energy lost $=\left(\frac{{\mathrm{E}}_1-{\mathrm{E}}_2}{{\mathrm{E}}_1}\right)\times 100$

$=\left(1-\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}\right)\times 100$

$=\left(1-\frac{{\mathrm{mgh}}_2}{{\mathrm{mgh}}_1}\right)\times 100$

$=\left(1-\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}\right)\times 100$

$=\left(1-\frac{2.4}{4}\right)\times 100$

$=\left(1-0.6\right)\times 100$

$=0.4\times 100=40\%$

Hence, percentage of energy lost is equal to $40\%$