Work, Energy and Power: Question 3

Worked solution

Part (a): Showing the increase in gravitational PE

The beam rises at constant speed, so all the useful work done goes into increasing its gravitational PE: $\Delta E_p = mg\Delta h = 250\times9.81\times8.0$

Working in two steps: $250\times9.81=2452.5$, then $2452.5\times8.0=19620\text{ J}$.

Check by recomputing the second step differently: $2452.5\times8.0=2452.5\times8=19620\text{ J}$ (since $2452.5\times4=9810$ and $9810\times2=19620$). Both routes agree.

So $\Delta E_p = 19620\text{ J} = 1.962\times10^{4}\text{ J}$, which rounds to $\boxed{1.96\times10^{4}}\text{ J}$ (3 s.f.), as required to show.

Part (b): Useful output power

Power is the rate of doing (useful) work, $P=\dfrac{W}{t}$. The useful work done by the motor equals the gain in gravitational PE found in part (a): $P_{\text{useful}} = \frac{\Delta E_p}{t} = \frac{19620}{20} = 981\text{ W}$

Check: $981\times20=19620\text{ J}$, which matches $\Delta E_p$ from part (a). Confirmed.

Part (c): Total power input

Efficiency is the ratio of useful power output to total power input: $\text{efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}}$

Rearranging for the total input power, using efficiency $=0.75$ (not $75$): $P_{\text{input}} = \frac{P_{\text{useful}}}{\text{efficiency}} = \frac{981}{0.75} = 1308\text{ W}$

Check: $1308\times0.75 = 981\text{ W}$, matching $P_{\text{useful}}$ from part (b). Confirmed. Note that $P_{\text{input}}>P_{\text{useful}}$, as expected for any real, non-ideal motor.

Part (d): Total energy supplied

The total input energy is the total input power multiplied by the time for which it acts: $E_{\text{input}} = P_{\text{input}}\times t = 1308\times20 = 26160\text{ J}$

Check using an independent route: the total input energy must also equal the useful energy output divided by the efficiency, $E_{\text{input}}=\Delta E_p/\text{efficiency}=19620/0.75=26160\text{ J}$. Both routes agree exactly, confirming the answer.

So $E_{\text{input}} = \boxed{26160}\text{ J} \approx 2.62\times10^{4}\text{ J}$ (3 s.f.).

Final answers

• (a) $\Delta E_p = \boxed{19620}\text{ J} \approx 1.96\times10^{4}\text{ J}$ (shown)
• (b) Useful output power $= \boxed{981}\text{ W}$
• (c) Total input power $= \boxed{1308}\text{ W}$
• (d) Total input energy $= \boxed{26160}\text{ J} \approx 2.62\times10^{4}\text{ J}$