Question

If a `5 kg` object is being constantly accelerated from `0m/s` to `12 m/s` over `6s`, how much power is being applied at `t= 3`?

Answer 1

`P = 30` W

Explanation:

Variables:
`m = 5` kg
`v_i = 0` m/s
`v_f = 12` m/s
`t = 6` s

Power is measured in Watts. A Watt is equivalent to energy over time.

Initial Energy:
`E(t=0) = 1/2 m v_i^2`
`E(t=0) = 1/2 (5) (0)^2 = 0` J

Final Energy:
`E(t=6) = 1/2 m v_f^2`
`E(t=6) = 1/2 (5) (12)^2`
`E(t=6) = 360` J

Power over the interval:
`P(t=6) = (E(t=6))/6 = 360/6 = 60` W

Since the acceleration is uniform, we can assume that the power was increasing uniformly during this period. So each second, the power increased by:
`P = (P(t=6))/6 = 60/6 = 10` W/s

So at `t=3`, the power would have been:
`P(t=3) = 3P = 3(10) = 30` W