A machine which is \(70 \%\) efficient raises a \(10 \mathrm{~kg}\) body through a certain distance and…
A machine which is \(70 \%\) efficient raises a \(10 \mathrm{~kg}\) body through a certain distance and spends \(100 \mathrm{~J}\) energy. The body is then released. On reaching the ground, the kinetic energy of the body will be
0
\(70 \mathrm{~J}\)
\(50 \mathrm{~J}\)
\(35 \mathrm{~J}\)
Solution
Energy spent by the machine to raise the body of mass \(10 \mathrm{~kg}\) at certain height, is \(E=100 \mathrm{~J}\)
Since, machine is only \(70 \%\) efficient.
Hence, energy used by machine,
\(\begin{aligned}
E^{\prime} & =70 \% \text { of } E \\
& =\frac{70}{100} \times 100=70 \mathrm{~J}
\end{aligned}\)
When body is released, then whole energy spent by machine is converted into kinetic energy on reaching the ground.
\(\therefore\) Kinetic energy, \(K=E^{\prime}=70 \mathrm{~J}\)