The total power divided by the external agent is mainly consumed in two parts:

(1) To move the body in the forward direction.
(2) To compensate the effect of the friction.
(i) Work done in moving the body
W₁ = F . S
W₁ = ma S …(1)
(ii) Work done agent friction
W₂ = f_k × S = ${\mu }_k$ R S = ${\mu }_k$ W S
W₂ = ${\mu }_k$ mg S …(2)
Net work done W = W₁ + W₂
W = ma S + ${\mu }_k$ mg S
W = m (a + $\mu$ g) S
Now, according to 2^nd equation of motion
S = u + $\frac{1}{2}$ at²
Now the power divided:–
$P=\frac{dw}{dt}=\frac{d}{dt}m\left(a+\mu g\right)\frac{1}{2}a{t}^2$
P = m(a + $\mu$ g) at