Work,Power and Energy

Assume a constant force $F$ is applied to push the box.

$F_{total} = F - \mu m g = m a \\ a = \frac{F - \mu m g}{m}$

Given that we reach velocity $v$ , find the time taken to get there:

$t_f = \frac{v}{a} = \frac{mv}{F - \mu m g}$

Determine the distance traveled:

$D = \frac{1}{2} a t_f^2 = \frac{1}{2} \frac{F - \mu m g}{m} \frac{m^2 v^2}{(F - \mu m g)^2} = \frac{m v^2}{2(F - \mu m g)}$

Determine the energy associated with the friction:

$E = \mu m g D = \mu m g \frac{m v^2}{2(F - \mu m g)} = \frac{\mu m^2 v^2 g}{2(F - \mu m g)}$

Calculate the average friction power:

$P = \frac{E}{t_f} = \frac{\mu m^2 v^2 g}{2(F - \mu m g)} \frac{F - \mu m g}{mv} = \large{\boxed{\frac{\mu m v g}{2}}}$

@Beatriz Sampaio , @Steven Chase , @Pi Han Goh , @Ken Kai , @Hana Nakkache , how did u solve this problem ? Could u pls post the solution ?