A machine carries a 4.0 kgpackage from an initial position of$\overrightarrow{{\mathbf{d}}_{\mathbf{i}}}\mathbf{=}\left(0.50m\right)\hat{\mathbf{i}}\mathbf{+}\left(0.75m\right)\hat{\mathbf{j}}\mathbf{+}\left(0.20m\right)\hat{\mathbf{k}}$at t=0to a final position of$\overrightarrow{{\mathbf{d}}_{\mathbf{f}}}\mathbf{=}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{k}}$at t=12 s. The constant force applied by the machine on the package is$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$. For that displacement, find (a) the work done on the package by the machine’s force and (b) the average power of the machine’s force on the package.

$$\begin{gathered} \overrightarrow{{\mathrm{d}}_{\mathrm{i}}}=(0.50\mathrm{m})\hat{\mathrm{i}}+(0.75\mathrm{m})\hat{\mathrm{j}}+(0.20\mathrm{m})\hat{\mathrm{k}}\mathrm{at}\mathrm{t}=0 \\ \overrightarrow{{\mathrm{d}}_{\mathrm{f}}}=(7.50\mathrm{m})\hat{\mathrm{i}}+(12.0\mathrm{m})\hat{\mathrm{j}}+(7.20\mathrm{m})\hat{\mathrm{k}}\mathrm{at}\mathrm{t}=12\mathrm{s} \\ \overrightarrow{\mathrm{F}}=(2.00\mathrm{N})\hat{\mathrm{i}}+(4.00\mathrm{N})\hat{\mathrm{j}}+(6.00\mathrm{N})\hat{\mathrm{k}} \end{gathered}$$

The work done on a particle by a constant force during its displacement is given as$\mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}$

Only the component of the force that is along the displacement can dothework on the object. The power due to force istherate at whichtheforce does the work on the object.

Formula:

$\mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}$

The dot product,

$$\begin{gathered} \left[a\hat{i}+b\hat{j}+c\hat{k}\right]\mathbf{.}\left[p\hat{i}+q\hat{j}+r\hat{k}\right]\mathbf{=}\left(\mathrm{ap}\right)\mathbf{+}\left(\mathrm{bq}\right)\mathbf{+}\left(\mathrm{cr}\right) \\ {\mathbf{P}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{W}}{\mathbf{∆}\mathbf{t}} \end{gathered}$$

The work done on a particle by a constant force during its displacement is given as

$\mathrm{W}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}}$

First, we determine the displacement

$\begin{array}{rcl}\overrightarrow{\mathrm{d}}& =& \overrightarrow{{\mathrm{d}}_{\mathrm{f}}}-\overrightarrow{{\mathrm{d}}_{\mathrm{i}}}\\ & =& (7.00\mathrm{m})\hat{\mathrm{i}}+(11.25\mathrm{m})\hat{\mathrm{j}}+(7.00\mathrm{m})\hat{\mathrm{k}}\end{array}$

Then,

$\begin{array}{rcl}\mathrm{W}& =& \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}}\\ \mathrm{W}& =& \left[(2.00\mathrm{N})\hat{\mathrm{i}}+(4.00\mathrm{N})\hat{\mathrm{j}}+(6.00\mathrm{N})\hat{\mathrm{k}}\right].\left[(7.00\mathrm{m})\hat{\mathrm{i}}+(12.00\mathrm{m})\hat{\mathrm{j}}+(7.25\mathrm{m})\hat{\mathrm{k}}\right]\\ \mathrm{W}& =& 14\mathrm{J}+45\mathrm{J}+42\mathrm{J}\\ & =& 101\mathrm{J}\\ & =& 1.01\times {10}^2\mathrm{J}\end{array}$

Hence the work done by the machine’s force on the package is $1.01\times {10}^2\mathrm{J}$.

The power due to force istherate at whichtheforce does the work on the object.

${\mathrm{P}}_{\mathrm{avg}}=\frac{\mathrm{W}}{∆\mathrm{t}}$

Here

$\begin{array}{rcl}∆\mathrm{t}& =& {\mathrm{t}}_2-{\mathrm{t}}_1\\ & =& 12\mathrm{s}\end{array}$

So,

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& \frac{\mathrm{W}}{∆\mathrm{t}}\\ & =& \frac{1.0\times {10}^2\mathrm{J}}{12\mathrm{s}}\\ & =& 8.4\mathrm{W}\end{array}$

The work done by the machine’s force on the package is $1.0\times {10}^2\mathrm{J}$and the average power of the machine’s force on the package is $8.4\mathrm{W}$.