A constant force of $5 \mathrm{N}$ accelerates a stationary particle of mass $500 \mathrm{gm}$ through a displacement of $5\mathrm{m}.$The average power delivered is

Given, force, $F=5 \mathrm{N}$

Mass, $m=500 \mathrm{g}=5\times {10}^{-1} \mathrm{kg}$

Displacement, $s=5 \mathrm{m}$

Work done is given by $W=\overrightarrow{F}·\overrightarrow{s}=Fs\cos \left(\theta \right)$

Let's assume the angle between force and displacement to be zero. Then work will be,

$W=Fs=5\times 5=25 \mathrm{J}$

Let the time taken to travel the displacement of $5 \mathrm{m}$ be $t \mathrm{s}$.
Applying Newton's second law of motion,

$F=ma$

$$\begin{gathered} \Rightarrow a=\frac{5}{5\times {10}^{-1}} \\ \Rightarrow a=10 \mathrm{m} {\mathrm{s}}^{-2} \end{gathered}$$

From the third equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow 5=0+\frac{1}{2}\times 10t^2$,           $\left(\text{initial velocity,} u=0\right)$

$\Rightarrow t=1 \mathrm{s}$

The average power is given by

$P_{\mathrm{avg}}=\frac{\text{Total work}}{\text{total time taken}}=\frac{25}{1}=25 \mathrm{W}$