Question

A machine which is $75\%$ efficient, uses $12\, J$ of energy in lifting $1 \,kg$ mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is

• A. $\sqrt{12} m / s$
• B. $\sqrt{18} m / s$
• C. $\sqrt{24} m / s$
• D. $\sqrt{32} m / s$

Answer 1

Answer: (B)

$\sqrt{18} m / s$

Answer 2

From law of conservation of energy
Potential energy $=$ kinetic energy
Given, potential energy $=\frac{75}{100} \times 12=9 J \ldots$. (i)
Kinetic energy $=\frac{1}{2} m v^{2}$...(ii)
Equating Eqs. (i) and (ii), we get
$\frac{1}{2} m v^{2}=9$
$\Rightarrow v=\sqrt{\frac{9 \times 2}{1}}=\sqrt{18} m / s$