A Machine Which Is 75 Percent Efficient Uses 12j Of Energy In Lifting Up A 1kg 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?

From the work-energy theorem,

Work done = change in potential energy

Work done = mgh

The potential energy of a body = 75% of 12J

mgh = 9J

\( h = \frac{9}{1*10} \)

= 0.9m

Now when this mass allows falling, then it acquires a velocity.

\( v = \sqrt{2gh} \) \( \Rightarrow \sqrt{2 * 10 * 0.9} \) \( \Rightarrow \sqrt{18}m/s \)

The velocity at the end of its fall is \( \sqrt{18}m/s \)

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