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 $ms^{- 1}$)

Question

Dr. Abubacker · Accepted Answer

(c) Energy stored $(E) = \frac{75}{100} 	imes (12) = 9\ J$
As $E = \frac{1}{2}mv^{2}$
$	herefore\ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 	imes 9}{1}} = \sqrt{18}\ ms^{- 1}$

Q. 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 $ms^{- 1}$)

Explanation

Reviewed by Dr. Abubacker

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