Answer
$\overline P=344W$
Work Step by Step
We have $$\overline P=Fv_{avg}$$
According to Newton's 2nd law of motion, $F=ma$
Acceleration $a$ can be written as: $a=\frac{v_f-v_0}{t}$
We also know for a constant $a$, $v_{avg}=\frac{v_0+v_f}{2}$
Therefore, $$\overline P=\frac{m(v_f-v_0)(v_f+v_0)}{2t}=\frac{m(v_f^2-v_0^2)}{2t} (1)$$
a) In the first case, we have $v_0=0$, $v_f=v$ and time $t$
$$\overline P_1=\frac{mv^2}{2t}=43W$$
b) In the second case, we have $v_0=0$, $v_f=2v$ and time $1/2t$
$$\overline P_2=\frac{m(4v^2)}{2(1/2t)}=8\frac{mv^2}{2t}=8\overline P_1=344W$$
