In three situations, a single force acts on a moving particle. Here are the velocities (at that instant) and the forces: (1) $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{}\mathbf{=}\left(-4\stackrel{⏜}{i}\right)\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{,}\mathbf{}\overline{\mathbf{f}}\mathbf{}\mathbf{=}\left(6\stackrel{⏜}{i}-20\right)\mathbf{N}\mathbf{:}$(2) $\stackrel{\mathbf{⏜}}{\mathbf{v}}\mathbf{=}\left(-3\stackrel{⏜}{i}+\stackrel{⏜}{j}\right)\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{,}\mathbf{}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{f}}\mathbf{=}\left(2\stackrel{⏜}{i}+6\stackrel{⏜}{j}\right)\mathbf{N}$(3) $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\left(-3\stackrel{⏜}{i}+\stackrel{⏜}{j}\right)\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{f}}\mathbf{=}\left(2\stackrel{⏜}{i}+6\stackrel{⏜}{j}\right)\mathbf{N}$. Rank the situations according to the rate at which energy is being transferred, greatest transfer to the particle ranked first, greatest transfer from the particle ranked last.

The three situations are:

$$\begin{gathered} \stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{-}\mathbf{4}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{F}\mathbf{=}\mathbf{6}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{-}\mathbf{20}\stackrel{\mathbf{⏜}}{\mathbf{j}} \\ \stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{2}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{-}\mathbf{3}\stackrel{\mathbf{⏜}}{\mathbf{j}\mathbf{}}\mathbf{\&}\mathbf{}\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{2}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{+}\mathbf{7}\stackrel{\mathbf{⏜}}{\mathbf{k}} \\ \stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{-}\mathbf{3}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{\&}\mathbf{}\mathbf{F}\mathbf{=}\mathbf{2}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{6}\stackrel{\mathbf{⏜}}{\mathbf{j}} \end{gathered}$$

Using the formula for power, i.e., rate of work done per unit time, we can rank the situations. So, using the formula of work in power, we can get the exact value of all the given situations.

Formulae:

The rate of energy transfer or work done of a body,$\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}$ (i)

The net energy transfer is equal to the work done by a body,$\mathrm{W}=\mathrm{F}.\mathrm{d}$ (ii)

We know that the power is the rate of energy transfer and is given using equation (ii) in equation (i) as follows:

$$\begin{gathered} \mathrm{P}=\mathrm{F}.\frac{\mathrm{d}}{\mathrm{t}} \\ =\mathrm{F}.\mathrm{v}..............\left(\mathrm{a}\right) \end{gathered}$$

So for case 1,

$\stackrel{\rightharpoonup }{\mathrm{v}}=-4\stackrel{⏜}{\mathrm{i}}\&\mathrm{F}=6\stackrel{⏜}{\mathrm{i}}-20\stackrel{⏜}{\mathrm{j}}$

So, the value of the power is given using the given values equation (a) as:

$$\begin{gathered} {\mathrm{P}}_1=\left(-4\stackrel{⏜}{\mathrm{i}}\right)\left(6\stackrel{⏜}{\mathrm{i}}-20\stackrel{⏜}{\mathrm{j}}\right) \\ =-24\mathrm{W} \end{gathered}$$

For case 2,

role="math" localid="1657171186237" $\stackrel{\rightharpoonup }{\mathrm{v}}=-2\stackrel{⏜}{\mathrm{i}}+3\stackrel{⏜}{\mathrm{j}}\&\mathrm{F}=-2\stackrel{⏜}{\mathrm{i}}+7\stackrel{⏜}{\mathrm{k}}$

So, the value of the power is given using the given values equation (a) as:

$$\begin{gathered} {\mathrm{P}}_2=\left(2\stackrel{⏜}{\mathrm{i}}-3\stackrel{⏜}{\mathrm{j}}\right).\left(-2\stackrel{⏜}{\mathrm{j}}+7\stackrel{⏜}{\mathrm{k}}\right) \\ =6\mathrm{W} \end{gathered}$$

For case 3,

$\stackrel{\rightharpoonup }{\mathrm{v}}=-3\stackrel{⏜}{\mathrm{i}}+\stackrel{⏜}{\mathrm{j}}\&\mathrm{F}=2\stackrel{⏜}{\mathrm{i}}+6\stackrel{⏜}{\mathrm{j}}$

So, the value of the power is given using the given values equation (a) as:

$$\begin{gathered} {\mathrm{P}}_3=\left(3\stackrel{⏜}{\mathrm{i}}+\stackrel{⏜}{\mathrm{j}}\right).\left(2\stackrel{⏜}{\mathrm{i}}+6\stackrel{⏜}{\mathrm{j}}\right) \\ =0\mathrm{W} \end{gathered}$$

So, the rank of the situations according to the rate of energy is 2 > 3 > 1.