91Ó°ÊÓ

The object is acted on by the force,

$\stackrel{→}{\mathrm{F}}=(4.0\text{N})−(2.0\text{N})+\left(9.0\text{N}\right)$

The object’s velocity is,

$\stackrel{→}{\mathrm{v}}=−(2.0\text{ }\text{m/s})+(4.0\text{ }\text{m/s})$

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

By takingthedot product of force$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{}$ and velocity$\stackrel{\mathbf{→}}{\mathbf{v}}$ vectors, the instantaneous rate at which the force does work on the object and for other instantaneous power$\mathbf{P}$,the velocity$\mathbf{V}$ of the object can be calculated.

Formula is as follow:

The instantaneous power is,$P=\stackrel{→}{F}.\stackrel{→}{v}$

Where,

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

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

$\stackrel{→}{\mathrm{F}}=(4.0\text{N})\widehat{\mathrm{i}}−(2.0\text{N})\widehat{\mathrm{j}}+(9.0\text{N})\widehat{\mathrm{k}}$

And, the object’s velocity is,

$\stackrel{→}{\mathrm{v}}=−\left(2.0\text{m/s}\right)\widehat{\mathrm{i}}+(4.0\text{m/s})\widehat{\mathrm{k}}$

Now, the instantaneous power is,

$P=\stackrel{→}{F}.\stackrel{→}{v}$

$\mathrm{P}=[\left(4.0\right)\widehat{\mathrm{i}}−\left(2.0\right)\widehat{\mathrm{j}}+(9.0)\widehat{\mathrm{k}}].[−(2.0)\widehat{\mathrm{i}}+(4.0)\widehat{\mathrm{k}}]$

$\mathrm{P}=\{\left(4.0\right)\widehat{\mathrm{i}}.[−(2.0)\widehat{\mathrm{i}}+(4.0)\widehat{\mathrm{k}}]−\left(2.0\right)\widehat{\mathrm{j}}.[−(2.0)\widehat{\mathrm{i}}+(4.0)\widehat{\mathrm{k}}]+(9.0)\widehat{\mathrm{k}}.[−(2.0)\widehat{\mathrm{i}}+(4.0)\widehat{\mathrm{k}}]\}$

$P=\{[−8.0+0.0]−[0.0+0.0]+[0.0+36]\}$

$P=28\text{W}$

Hence, the instantaneous rate at which the force does work on the object is $P=28\text{W.}$

Similarly, instantaneous power is,

$\mathrm{P}=\stackrel{→}{\mathrm{F}}.\stackrel{→}{\mathrm{v}}$

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

$P=(−2.0)v$

Therefore, the velocity is,

$v=\frac{P}{(−2.0)}$

$v=\frac{−12}{(−2.0)}$

$v=6.0\text{ }\text{m/s}$

Hence, for instantaneous power $P=−12$, the velocity of the object is $v=6.0\text{ }\text{m/s.}$