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. The velocity at the end of its fall is (in `ms^(-1)`)

Explanation

Potential energy of the mass at a height above the earth's surface is given by `=(75)/(100)xx12=9 J`………(i)
Now, KE of the mass at the end of fall `KE=(1)/(2)mv^(2)`……….(ii)
Applying law of conservation of energy,
`(1)/(2)mv^(2)=9impliesv=sqrt((2xx9)/(m))sqrt((18)/(1))=sqrt(18) ms^(-1)`