Question

A quarter horse power motor runs at a speed of $600\, rpm$ . Assuming $40\%$ efficiency, the work done by the motor in the one rotation will be

• A. $7.46\,J$
• B. $74.6\,J$
• C. $7400\,J$
• D. $7.46\,erg$

Answer 1

Answer: (A)

$7.46\,J$

Answer 2

Given, power is $\frac{1}{4} nP = \frac{764}{4}W = 186.5\,W$
Since, efficiency of motor is $40\%$.
The power used in doing work is $40\%$ of $186.5W$, so
$P = 186.5 \times \frac{40}{100} = 74.6\,W$
Angular velocity of the motor, $\omega = 600\, rpm = 10 \,rps$
$\Rightarrow \omega = (600)(2\pi)60\,rad/s$
$\Rightarrow \omega = 20\,\pi \,rad /s$
Let the torque be $T$.
So, $P = T\omega$
$\Rightarrow T = \frac{P}{\omega} = \frac{74.6}{20\,\pi}$
Now, work done in $1$ rotation $= T(2n)$
$W = \frac{74.6}{20\,\pi} \times 2\pi$
$W = 7.46\,J$