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?

Respuesta :

Answer:

Approximately [tex]0.35\; \rm W[/tex] on average.

Explanation:

There are multiple to find power:

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

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

Therefore, the equation [tex]P = W / t[/tex] seem to be more suitable.

[tex]\begin{aligned}&P\\ &= \frac{W}{t} \\ &= \frac{0.70\; \rm J}{2\; \rm s} \approx 0.35\; \rm W\end{aligned}[/tex].