Question

(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?

Short answer

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

Step-by-step solution

Given information

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

Determining the concept

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.

(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,

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

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 $P=28W$

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

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 $v=6.0m/s$