91Ó°ÊÓ

It is given that,

Force acting on the object is,$\stackrel{→}{F}=(3.00\mathrm{N})\hat{\mathrm{i}}+(7.00\mathrm{N})\hat{\mathrm{j}}+(7.00\mathrm{N})\hat{\mathrm{k}}$

The mass of the mobile object is, m = 2.00 kg.

Initial position of the object is,${\stackrel{→}{d}}_i=(3.00\mathrm{m})\hat{\mathrm{i}}+(2.00\mathrm{m})\hat{\mathrm{j}}+(7.00\mathrm{m})\hat{\mathrm{k}}$

Final position of the object is,${\stackrel{→}{d}}_f=-(5.00\mathrm{m})\hat{\mathrm{i}}+(4.00\mathrm{m})\hat{\mathrm{j}}+(7.00\mathrm{m})\hat{\mathrm{k}}$

Time interval is, t = 4.00 s.

The problem deals with the power and the work done. Work,energy, and power are the fundamental concepts of physics. Work is the displacement of an object when force is applied on it. And the power is the rate at which the work is done.Here to initiate, the displacement$\stackrel{\mathbf{→}}{\mathbf{d}}$of the object can be found. Then, takingthedot product of role="math" localid="1657170094293" $\stackrel{\mathbf{→}}{\mathit{F}}$forceand displacementrole="math" localid="1657170082575" $\stackrel{\mathit{→}}{d}$vectors, the work doneon the object will be found. By takingtheratio of work doneW to interval t , find the average power due to the given force. By calculating the values of magnitudesrole="math" localid="1657170106850" ${\stackrel{\mathit{→}}{d}}_i$androle="math" localid="1657170117676" ${\stackrel{\mathit{→}}{d}}_f$also taking the dot product of andand using these values, calculate the angle between role="math" localid="1657170130919" ${\stackrel{\mathit{→}}{d}}_i$and role="math" localid="1657170141717" ${\stackrel{\mathit{→}}{d}}_f$.

Formulae:

The object displacement role="math" localid="1657170213868" $\stackrel{\mathit{→}}{d}$is,

$\stackrel{\mathit{→}}{d}={\stackrel{\mathit{→}}{d}}_f-{\stackrel{\mathit{→}}{d}}_i$ (i)

The work done is,

$W=\stackrel{→}{F}.\stackrel{→}{d}$ (ii)

The average power is,

$P_{avg}=\frac{W}{t}$ (iii)

The angle between vectors $\stackrel{→}{d_i}$and $\stackrel{→}{d_f}$is,

$\mathrm{ϕ}={\cos }^{-1}\left(\frac{\stackrel{→}{d_i}.\stackrel{→}{d_f}}{d_i{d}_f}\right)$ (iv)

Where,

$P_{avg}$is average power, $\stackrel{→}{F}$is force vector, $\stackrel{→}{d}$ is displacement vector, Wis the work done, t is time, is the angle between vectors ${\stackrel{→}{d}}_i$and ${\stackrel{→}{d}}_f$and ${\stackrel{→}{d}}_f,{\stackrel{→}{d}}_i,\stackrel{→}{d}$are final, initial and total displacement vectors.

The object displacement $\stackrel{→}{d}$is given as,

$$\begin{gathered} \stackrel{→}{d}={\stackrel{→}{d}}_f-{\stackrel{→}{d}}_i \\ \stackrel{→}{d}=\left[-(5.00\mathrm{m})\hat{i}+(7.00\mathrm{m})\hat{j}+(7.00\mathrm{m})\hat{k}\right]-\left[-(3.00\mathrm{m})\hat{i}-(2.00\mathrm{m})\hat{j}+(5.00\mathrm{m})\hat{k}\right] \\ \stackrel{→}{d}=\left[-(8.00\mathrm{m})\hat{i}+(6.00\mathrm{m})\hat{j}+(2.00\mathrm{m})\hat{k}\right] \end{gathered}$$

With equation (ii), the work done is given by,

role="math" localid="1657172387746" $$\begin{gathered} W=\stackrel{→}{F}.\stackrel{→}{d} \\ W=\left[(3.00\mathrm{N})\hat{i}+(7.00\mathrm{N})\hat{j}+(7.00\mathrm{N})\hat{k}\right]-\left[-(8.00\mathrm{m})\hat{i}+(6.00\mathrm{m})\hat{j}+(2.00\mathrm{m})\hat{k}\right] \\ W=\left\{\left[-24.0+0.00+0.00\right]+\left[0.00+42.0+0.00\right]+0.00+0.00+14.0\right\} \\ W=32.0\mathrm{J} \end{gathered}$$

Hence, the work done W on the object by the given force $\stackrel{→}{F}$in given time is W = 32.0 J.

Using equation (iii) the average power $P_{avg}$is given as,

$$\begin{gathered} P_{avg}=\frac{W}{t} \\ {P}_{avg}=\frac{32.0}{4.00} \\ {P}_{avg}=8.00\mathrm{W} \end{gathered}$$

Hence, the average power $P_{avg}$due to the forcerole="math" localid="1657171847531" $\stackrel{\mathit{→}}{F}$during given interval t is$P_{avg}=8.00\mathrm{W}$

The distance from the coordinate origin to the initial position is,

$$\begin{gathered} d_i=\sqrt{\left[{(3.00)}^2+{(-2.00)}^2+{(5.00)}^2\right]} \\ {d}_i=\sqrt{\left[9.00+4.00+25.0\right]} \\ {d}_i=6.16\mathrm{m} \end{gathered}$$

Similarly, the distance from the coordinate origin to the final position is,

$$\begin{gathered} d_f=\sqrt{\left[-{(5.00)}^2+{(4.00)}^2+{(7.00)}^2\right]} \\ {d}_f=\sqrt{\left[25.0+16.0+49.0\right]} \\ {d}_f=9.49\mathrm{m} \end{gathered}$$

Now, their dot product is,

$$\begin{gathered} {\stackrel{→}{d}}_i.\stackrel{→}{d_f}=\left[(3.00\mathrm{N})\hat{i}-(2.00\mathrm{N})\hat{j}+(5.00\mathrm{N})\hat{k}\right].\left[-(5.00\mathrm{m})\hat{i}+(4.00\mathrm{m})\hat{j}+(7.00\mathrm{m})\hat{k}\right] \\ {\stackrel{→}{d}}_i.\stackrel{→}{d_f}=\left[-15.0\mathrm{m}+0.00\mathrm{m}-0.00\mathrm{m}\right]+\left[0.00\mathrm{m}-8.00\mathrm{m}-0.00\mathrm{m}\right]+\left[0.00\mathrm{m}-0.00\mathrm{m}-35.0\mathrm{m}\right] \\ {\stackrel{→}{d}}_i.\stackrel{→}{d_f}=12.0\mathrm{m} \end{gathered}$$

It is known that,

${\stackrel{→}{d}}_i.\stackrel{→}{d_f}=d_i{d}_f\cos \left(\mathrm{ϕ}\right)$

Therefore, using equation (IV) the angle between vectors ${\stackrel{→}{d}}_i$and ${\stackrel{→}{d}}_f$is,

$$\begin{gathered} \mathrm{ϕ}={\cos }^{-1}\left(\frac{{\stackrel{→}{d}}_i.{\stackrel{→}{d}}_f}{d_i{d}_f}\right) \\ \mathrm{ϕ}={\cos }^{-1}\left(\frac{12.0}{6.16×9.49}\right) \\ \mathrm{ϕ}={\cos }^{-1}(0.205) \\ \mathrm{ϕ}=1.36\mathrm{rad} \\ =78.2^{°} \end{gathered}$$

Hence, the angle between vectors ${\stackrel{→}{d}}_i$and ${\stackrel{→}{d}}_f$is $\mathrm{ϕ}=1.36\mathrm{rad}=78.2^{°}$,

By calculating the displacementof the object, the work done on the object can be found. By using this value of work done and time interval, the average power due to the given force can be calculated. By calculating the values of magnitudes and the dot product ofthe given displacement vectors,the angle between them can be calculated.