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  4. A machine which is 70 % efficient raises a 10 kg body through a certain distance and spends 100 J of energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

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Q. A machine which is $70 \%$ efficient raises a $10\, kg$ body through a certain distance and spends $100\, J$ of energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

Work, Energy and Power

Solution:

$70 \%$ efficiency means body will get $70 \%$ of
energy spent by the machine $=\frac{70}{100} \times 100 J =70 J$
Hence P.E. of body $=70 J$ On reaching ground PE becomes $KE =70 J$.