(a) at a certain instant, a particle-like object is acted on by a force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\left(4.0N\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{-}\left(2.0N\right)\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{+}\left(9.0N\right)\stackrel{\mathbf{⏜}}{\mathbf{k}}$ while the object’s velocity is$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{}\mathbf{-}\left(2.0m/s\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\left(4.0m/s\right)\stackrel{\mathbf{⏜}}{\mathbf{k}}$ . What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a y component. If the force is unchanged and the instantaneous power is -12 W , what is the velocity of the object?

The object is acted on by the force,

$\stackrel{\rightharpoonup }{\mathrm{F}}=\left(4.0\mathrm{N}\right)\stackrel{⏜}{\mathrm{i}}-\left(2.0\mathrm{N}\right)\stackrel{⏜}{\mathrm{j}}+\left(9.0\mathrm{N}\right)\stackrel{⏜}{\mathrm{k}}$

The object’s velocity is,

$\stackrel{\rightharpoonup }{\mathrm{v}}=-\left(2.0\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{i}}+\left(4.0\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{k}}$

Instantaneous power $\mathrm{P}=12\mathrm{W}$

By takingthedot product of force and velocity vectors, the instantaneous rate at which the force does work on the object and for other instantaneous power,the velocity of the object can be calculated.

Formula is as follow:

The instantaneous power is,$\mathrm{P}={\stackrel{\rightharpoonup }{\mathrm{F}}}_{.\stackrel{\rightharpoonup }{\mathrm{v}}}$

Where,

P isinstantaneous power,$\stackrel{\rightharpoonup }{F}$is force vector and$\stackrel{\rightharpoonup }{v}$is velocity vector.

The object is acted on by the force is given as,

$\stackrel{\rightharpoonup }{\mathrm{F}}=\left(4.0\mathrm{N}\right)\stackrel{⏜}{\mathrm{i}}-\left(2.0\mathrm{N}\right)\stackrel{⏜}{\mathrm{j}}+\left(9.0\mathrm{N}\right)\stackrel{⏜}{\mathrm{k}}$

And, the object’s velocity is,

$\stackrel{\rightharpoonup }{\mathrm{v}}=-\left(2.0\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{i}}+\left(4.0\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{k}}$

Now, the instantaneous power is,

$\mathrm{P}={\stackrel{\rightharpoonup }{\mathrm{F}}}_{.\stackrel{\rightharpoonup }{\mathrm{v}}}$

$$\begin{gathered} P=\left[\left(4.0\right)\stackrel{⏜}{i}-\left(2.0\right)\stackrel{⏜}{j}+\left(9.0\right)\stackrel{⏜}{k}\right].\left[-\left(2.0\right)\stackrel{⏜}{i}+\left(4.0\right)\stackrel{⏜}{k}\right] \\ P=\left\{\left(4.0\right)\right.\stackrel{⏜}{i}.\left[-\left(2.0\right)\stackrel{⏜}{j}+\left(4.0\right)\stackrel{⏜}{k}\right]-\left(2.0\right)\stackrel{⏜}{j}.\left[-\left(2.0\right)\stackrel{⏜}{i}+\left(4.0\right)\stackrel{⏜}{k}\right]+\left(9.0\right)\stackrel{⏜}{k.}\left[-\left(2.0\right)\stackrel{⏜}{i}+\left(4.0\right)\stackrel{⏜}{k}\right]\left.\right\} \\ P=\left\{\left[-8.0+0.0\right]\right.-\left[0.0+0.0\right]+\left[0.0+36\right]\left.\right\} \\ P=28W \end{gathered}$$

Hence, the instantaneous rate at which the force does work on the object is $P=28W$

Similarly, instantaneous power is,

$P=\stackrel{\rightharpoonup }{F.}\stackrel{\rightharpoonup }{v}$

For instantaneous power $P=12W\stackrel{\rightharpoonup }{v}$ consists only y component, therefore,

$P=\left(-2.0\right)v$

Therefore, the velocity is,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{P}}{\left(-2.0\right)} \\ \mathrm{v}=\frac{-12}{\left(-2.0\right)} \\ \mathrm{v}=6.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, for instantaneous power $P=-12$, the velocity of the object is $v=6.0m/s$