Question

A does 500J of work in 10 minutes and B does 600J of work in 20 minutes. The power delivered by A and B is ${P_1}$ and ${P_2}$ respectively. Find the relation between ${P_1}$ and ${P_2}$.
A. ${P_1} = {P_2}$
B. ${P_1} > {P_2}$
C. ${P_1} < {P_2}$
D. ${P_1}$ and ${P_2}$ are undefined

Answer 1

The power is the work done per unit time. Calculate the work done by A and work done by B. The unit of power is joule per second. The power is the amount of energy transferred in unit time. It is a scalar quantity. The S.I unit of power is joule per second. The conventional unit for power is watt.

Formula used:
Power, $P = \dfrac{W}{t}$, where, W is the work done and t is the time.

Complete step by step answer:
We have given the work done by A, ${W_A} = 500\,{\text{J}}$ in time ${t_A} = 10\,{\text{min}} = 600\,{\text{sec}}$ and the work done by B, ${W_B} = 600\,{\text{J}}$ in time ${t_B} = 20\,{\text{min}} = 1200\,{\text{sec}}$.
We have the relation between power and work done. The power is the work done per unit time. Therefore, the power delivered by A is,
${P_1} = \dfrac{{{W_A}}}{{{t_A}}}$
Substituting ${W_A} = 500\,{\text{J}}$ and ${t_A} = 600\,{\text{sec}}$ in the above equation, we get,
${P_1} = \dfrac{{500}}{{600}} = 0.83\,{\text{J/s}}$
Let us express the power delivered by B as,
${P_2} = \dfrac{{{W_2}}}{{{t_2}}}$
Substituting ${W_B} = 600\,{\text{J}}$ and ${t_B} = 1200\,{\text{sec}}$ in the above equation, we get,
$\therefore{P_2} = \dfrac{{600}}{{1200}} = 0.5\,{\text{J/s}}$
Therefore, the relation between ${P_1}$ and ${P_2}$ is ${P_1} > {P_2}$.

So, the correct answer is option B.

Note: While using the formula for work done, make sure that you have converted all the quantities in the S.I unit. The S.I unit of time is second and not minute. The power is not a vector quantity since the work done is a scalar quantity.The work done is the amount of energy transferred or gained. It is the product of applied force and displacement of the body in the direction of the force. The S.I unit work done is joule.