Question

(a) at a certain instant, a particle-like object is acted on by a force $\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{)}\widehat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{)}\widehat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{)}\widehat{\mathbf{k}}$ while the object’s velocity is $\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\widehat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\widehat{\mathbf{k}}$. 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 $\mathbf{-}\mathbf{12}\mathit{W}$ , what is the velocity of the object?

Short answer

1. The instantaneous rate at which the force does work on the object is$\mathrm{P}=28\text{W.}$
2. For instantaneous power $\mathrm{P}=-12$, the velocity of the object is$\mathrm{v}=6.0\text{m/s.}$

Step-by-step solution

Given information

The object is acted on by the force,

$\overrightarrow{\mathrm{F}}=(4.0\text{N})-(2.0\text{N})+\left(9.0\text{N}\right)$

The object’s velocity is,

$\overrightarrow{\mathrm{v}}=-(2.0\text{m/s})+(4.0\text{m/s})$

Instantaneous power $\mathrm{P}=12\text{\hspace{0.17em}W}$

Determining the concept

By takingthedot product of force$\overrightarrow{\mathbf{F}}\mathbf{}$ and velocity$\overrightarrow{\mathbf{v}}$ 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,$P=\overrightarrow{F}.\overrightarrow{v}$

Where,

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

(a) Determining the instantaneous rate at which the force does work on the object 

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

$\overrightarrow{\mathrm{F}}=(4.0\text{N})\widehat{\mathrm{i}}-(2.0\text{N})\widehat{\mathrm{j}}+(9.0\text{N})\widehat{\mathrm{k}}$

And, the object’s velocity is,

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

Now, the instantaneous power is,

$P=\overrightarrow{F}.\overrightarrow{v}$

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

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

$P=\{[-8.0+0.0]-[0.0+0.0]+[0.0+36]\}$

$P=28\text{W}$

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

(b) determining the velocity of the object at some other time 

Similarly, instantaneous power is,

$\mathrm{P}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{v}}$

For instantaneous power $P=12\text{\hspace{0.17em}W}\overrightarrow{v}$consists only y component, therefore,

$P=(-2.0)v$

Therefore, the velocity is,

$v=\frac{P}{(-2.0)}$

$v=\frac{-12}{(-2.0)}$

$v=6.0\text{m/s}$

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