Question

If a `12 kg` object is constantly accelerated from `0m/s` to `40 m/s` over 8 s, how much power musy be applied at `t=1`?

Answer 1

`300"W"` (Watts)

Explanation:

The power applied to an object is defined as `P=W/t`, and expanding the definition of work, `P=(Fd)/t` (assuming force and displacement are in the same direction). `d/t=v`, so `P=Fv`. `F=ma`, so `P=mav`.

The acceleration of an object is `(vec(v_f)-vec(v_i))/(Deltat)`, for this object it is `(40"m/s"-0"m/s")/(8"s")=5"m/s"^2`.

The velocity of an object at time `t` is `at`, so in this case the velocity at `1"s"` is `5"m/s"^2*1"s"=5"m/s"`

Finally, solving for power using `P=mav`, we get `P=12"kg"*5"m/s"^2*5"m/s"=300"W"`