Question

A machine, which is 75 percent efficient uses 12 joules of energy in lifting up a 1 kg 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. $\sqrt{32}m/s$
• B. $\sqrt{24}m/s$
• C. $\sqrt{18}m/s$
• D. $\sqrt{9}m/s$

Answer 1

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

given efficiency of the machine is 75% and it is spending 12 J of energy to lift that mass
according to the law of conservation of energy the stored potential energy in the mass will be converted to its kinetic energy. hence,
$(12\times 75)/100=(m\times v^2)/2$
$V=\sqrt{18}$ $m/s$