A machine which is 75% efficient, uses 12 , J | vedclass

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Question

(C) The energy supplied by the machine is $E_{in} = 12 \, J$. Since the efficiency is $75\%$, the useful work done $(W)$ in lifting the mass is $W = 0

Vedclass · Accepted Answer

$\sqrt{18}$

Vedclass · Answer

(C) The energy supplied by the machine is $E_{in} = 12 \, J$. Since the efficiency is $75\%$, the useful work done $(W)$ in lifting the mass is $W = 0.75 	imes 12 \, J = 9 \, J$. This work is stored as gravitational potential energy: $mgh = 9 \, J$. Given $m = 1 \, kg$ and taking $g = 10 \, m \, s^{-2}$, we have $1 	imes 10 	imes h = 9$, which gives $h = 0.9 \, m$. When the mass falls through this distance $h$, its potential energy is converted into kinetic energy. Using the conservation of energy, $\frac{1}{2}mv^2 = mgh$. $v = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 0.9} = \sqrt{18} \, m \, s^{-1}$.

$A$ machine which is $75\%$ efficient, uses $12 \, J$ of energy in lifting up a $1 \, kg$ mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in $m \, s^{-1}$):

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