Power — Question 2

Question 2

A signed power history separates gross delivery from net work

A device delivers mechanical power that rises linearly from 0 to 100 W during 0–2 s, remains at 100 W during 2–4 s, and is $-50$ W during 4–6 s. Transitions at 4 s are idealized finite steps.

1. Find positive work delivered, energy received back, and net work.

2. Find average signed power over the full six seconds.

3. If all positive mechanical output requires 80 percent conversion efficiency and no returned energy is recovered electrically, find the required electrical input energy.

Question 2 - Solution

A signed power history separates gross delivery from net work

(a) Add signed areas.

The governing relation is $$\boxed{W=\int_{t_i}^{t_f}P(t)\,dt,\qquad\overline P=\frac{W}{t_f-t_i}}$$ Positive work is $$\tfrac12(2)(100)+(2)(100)=300\ \mathrm J.$$ The final interval returns 100 J mechanically to the device. Net work is $$300-100=200\ \mathrm J.$$

(b) Divide net work by total elapsed time.

$$\overline P=200/6=100/3\ \mathrm W.$$ This is not the average of the largest and smallest power values.

(c) Apply efficiency to the gross output.

Electrical input is $$300/.80=375\ \mathrm J.$$ The unrecovered return does not offset that input. Using $200/.80$ would incorrectly credit returned energy as reusable electrical energy.