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. $\sqrt{12} \, ms^{ - 1}$
• B. $\sqrt{18} \, ms^{ - 1}$
• C. $\sqrt{24} \, ms^{ - 1}$
• D. $\sqrt{32} \, ms^{ - 1}$

Answer 1

Answer: (B)

$\sqrt{18} \, ms^{ - 1}$

Answer 2

Potential energy of the mass at a height above the surface is given by
$= \frac{ 75 }{ 100 } \times 12 = 9$ J \hspace20mm ..(i)
Now, KE of the mass at the end of fall
KE = $\frac{1}{2} mv^2$ \hspace20mm ..(ii)
Applying law of conservation of energy
$\frac{1}{2} mv^2 = 9$
v = $\sqrt{ \frac{ 2 \times 9 }{ m }}$
$\sqrt{ \frac{ 18 }{ 1}} = \sqrt{18 } \, ms^{ - 1}$