Solve the following :
Question:

A constant force of $2.50 \mathrm{~N}$ accelerates a stationary particle of mass $15 \mathrm{~g}$ through a displacement of $2.50 \mathrm{~m}$. Find the work done and the average power delivered?

Solution:

$W=F \cdot d$

$W=F d \cos \theta$

$W=2.5 \times 2.5 \times \cos \theta$

$W=6.25 \mathrm{~J}$

Now,

$W=\frac{1}{2} m v^{2}-\frac{1}{2} m v^{2}$

$\frac{1}{2} m v^{2}-0=6.25$

$v=28.8 \mathrm{~m} / \mathrm{s}$

And $F=m a$

$\mathrm{a}=\frac{F}{m}=2.5 / 0.015=166.66 \mathrm{~m} / \mathrm{s}^{2}$

So, v $=u+a t$

$t=\frac{(v-u)}{a}=28.8 / 166.66=0.1728 \mathrm{sec}$

And average power, $\mathrm{P}_{\mathrm{w}}=\frac{}{\frac{w}{t}}=6.25 / 0.1728=36.1 \mathrm{~W}$