Question

A chipmunk (mass of approximately 1 kg) does push-ups by using a force of 5 N to elevate its center-of-mass by 6 cm in order to do 0.70 Joule of work. If the chipmunk does all this work in 2 seconds, what is its power?

Answer 1

Approximately
`0.35\; \rm W` on average.

Step-by-step explanation:

There are multiple to find power:

• Power
  `P` is equal to the product of force
  `F` and the speed
  `v` at which that force moves the object. That is:
  `P = F \cdot v`.
• On the other hand, average power
  `P` is also equal to the rate at which work
  `W` is done. In other words, average power
  `P\!` is the average work done in unit time. If work
  `W\!` is done in time
  `t`, the average power would be
  `P = W / t`.

The question did not state the speed at which the arm of the chipmunk moves. However, the question did state the work done (
`W = 0.70\; \rm J`) and the time required (
`t = 2\; \rm s`).

Therefore, the equation
`P = W / t` seem to be more suitable.

`\begin{aligned}&P\\ &= (W)/(t) \\ &= (0.70\; \rm J)/(2\; \rm s) \approx 0.35\; \rm W\end{aligned}`.