Question

A lifting machine, having an efficiency of 80 % uses 2500 J of energy in lifting a 10 kg load over a certain height. If the load is now allowed to fall through that height freely, its velocity at the end of the fall will be
(take acceleration due to gravity as $10m/s^2$ )

• A. $10m{s}^{-1}$
• B. $15m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $25m{s}^{-1}$

Answer 1

The correct option is C $20m{s}^{-1}$

1) Energy stored in the form of $P.E=0.8\times 2500J$
2)This energy will again be converted into K.E
i.e $\frac{1}{2}m{v}^2=0.8\times 2500$
$\frac{1}{2}\times 10\times v^2=0.8\times 2500$
$v^2=400$
$v=20$ m/s