Question

A machine which is 75 percent efficient uses $12J$ of energy in lifting up a $1kg$ mass through a certain distance. The mass is then allowed to fall through that distance. What will its velocity be at the end of its fall?

• A. $\sqrt{18}m/s$
• B. $\sqrt{36}m/s$
• C. $\sqrt{8}m/s$
• D. $\sqrt{9}$

Answer 1

The correct option is A $\sqrt{18}m/s$

From work energy theorem,

Work done $=$ change in potential energy

So, $W=mgh$

So, $0.75\times 12=1\times 10\times h$

So, $h=0.9m$

Now, from conservation of energy,

$KE=PE$

So, $\frac{1}{2}m{v}^2=mgh$

So, $v=\sqrt{2gh}=\sqrt{2\times 10\times 0.9}=3\sqrt{2}m/s$