A force $\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{9}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{30}\hat{\mathbf{k}}\mathbf{)}\mathbf{N}$acts on a 2.90 kg object that moves in time interval 2.10 s from an initial position ${\overrightarrow{\mathbf{r}}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{7}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{90}\hat{\mathbf{j}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{50}\hat{\mathbf{k}}\mathbf{)}\mathbf{m}$to a final position ${\overrightarrow{\mathbf{r}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{10}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{30}\hat{\mathbf{j}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{40}\hat{\mathbf{k}}\mathbf{)}\mathbf{m}$. Find (a) the work done on the object by the force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between vectors ${\overrightarrow{\mathbf{r}}}_{\mathbf{1}}$and ${\overrightarrow{\mathbf{r}}}_{\mathbf{2}}$.

It is given that,

Force is$\overrightarrow{F}=(2.00\hat{i}+9.00\hat{j}+5.30\hat{k})$.

Initial position islocalid="1657947460948" ${\overrightarrow{r}}_1=2.70\hat{\mathrm{i}}-2.90\hat{\mathrm{j}}+5.50\hat{\mathrm{k}}$.

Final position is ${\overrightarrow{r}}_2=4.10\hat{i}-3.30\hat{j}+5.40\hat{k}$.

Mass is m= 2.90 kg.

Time is t= 2.10 s

The problem deals with the work done and power which are the fundamental concept of physics.Work is the displacement of an object when force is applied on it.Find distance d by calculating the difference between the given vectors, and then find the work done. Using the work done found in part (a), find the power for the given time interval. Use the dot product of vectors to find the angle between them.

Formulae:

Work done is given by,

$$\begin{gathered} W=Fd \\ P=\frac{W}{t} \\ {\overrightarrow{r}}_1.{\overrightarrow{r}}_2=r_1{r}_2\cos \theta \end{gathered}$$

Where,

F istheforce, P isthepower, t isthetime, d is the displacement, $\theta$is the angle between the position vectors and $r_1,r_2$are the position vectors.

First, find the distance d from the difference between the given vectors,

$$\begin{gathered} d={\overrightarrow{r}}_2-{\overrightarrow{r}}_1 \\ d=\left(-4.10\hat{\mathrm{i}}+3.30\hat{\mathrm{j}}+5.40\hat{\mathrm{k}}\right)-\left(2.70\hat{\mathrm{i}}-2.90\hat{\mathrm{j}}+5.50\hat{\mathrm{k}}\right) \\ d=\left(-6.80\hat{\mathrm{i}}+6.20\hat{\mathrm{j}}+0.10\hat{\mathrm{k}}\right) \end{gathered}$$

Now, the work done is,

$$\begin{gathered} W=Fd \\ W=\left(2.00\hat{\mathrm{i}}+9.00\hat{\mathrm{j}}+5.30\hat{\mathrm{k}}\right)-\left(-6.80\hat{\mathrm{i}}+6.20\hat{\mathrm{j}}+0.10\hat{\mathrm{k}}\right) \\ W=-13.6+55.8-0.53 \\ W=41.7\mathrm{J} \end{gathered}$$

Hence, work done on the object by the force is 41.7 J

Work done and time have been found, now power is,

$$\begin{gathered} P=\frac{W}{t} \\ P=\frac{41.67}{2.10} \\ =19.84\mathrm{W} \\ P=19.8\mathrm{W} \end{gathered}$$

Hence, the average power due to the force in the given time interval is P = 19.8 W

Calculate the dot product of the given vectors to find the angle between them,

$$\begin{gathered} {\overrightarrow{r}}_1.{\overrightarrow{r}}_2=r_1{r}_2\cos \theta \\ \cos \theta =\frac{{\overrightarrow{r}}_1.{\overrightarrow{r}}_2}{r_1{r}_2} \\ \cos \theta =\frac{9.06}{51.1147} \\ \theta =\left(0.1772\right) \\ \theta =79.79 \\ =79.8^{°} \end{gathered}$$

Hence, the angle between the vectors ${\overrightarrow{r}}_1\mathrm{and}{\overrightarrow{r}}_2\mathrm{is}\theta =79.8^{°}$