Question

A force $\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$acts on a 2.00 kg mobile object that moves from an initial position of role="math" localid="1657168721297" $\overrightarrow{{\mathbf{d}}_{\mathbf{i}}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$to a final position of $\overrightarrow{{\mathbf{d}}_f}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$in 4.00 s. Find (a) the work done on the object by the force in the 4.00 sinterval, (b) the average power due to the force during that interval, and (c) the angle between vectors role="math" localid="1657168815303" ${\overrightarrow{\mathit{d}}}_{\mathit{i}}$and${\overrightarrow{\mathit{d}}}_{\mathit{f}}$.

Short answer

1. The work done Won the object by the given force $\overrightarrow{F}$in given time is W = 32.0 J
2. The average power $P_{avg}$due to the force during given interval t is $P_{avg}=8.00\mathrm{W}$
3. The angle between vectors ${\overrightarrow{d}}_i$and ${\overrightarrow{d}}_f$is,$\varphi =1.36\mathrm{rad}=78.2^{°}$

Step-by-step solution

Given information

It is given that,

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

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

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

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

Time interval is, t = 4.00 s.

Determining the concept

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$\overrightarrow{\mathbf{d}}$of the object can be found. Then, takingthedot product of role="math" localid="1657170094293" $\overrightarrow{\mathit{F}}$forceand displacementrole="math" localid="1657170082575" $\overrightarrow{d}$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" ${\overrightarrow{d}}_i$androle="math" localid="1657170117676" ${\overrightarrow{d}}_f$also taking the dot product of andand using these values, calculate the angle between role="math" localid="1657170130919" ${\overrightarrow{d}}_i$and role="math" localid="1657170141717" ${\overrightarrow{d}}_f$.

Formulae:

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

$\overrightarrow{d}={\overrightarrow{d}}_f-{\overrightarrow{d}}_i$ (i)

The work done is,

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

The average power is,

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

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

$\varphi ={\cos }^{-1}\left(\frac{\overrightarrow{d_i}.\overrightarrow{d_f}}{d_i{d}_f}\right)$ (iv)

Where,

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

(a) Determining the work done W on the object by the given force in the given time

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

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

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

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

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

(b) Determining the average power Pavgdue to the force during F→given interval t is Pavg=8.00W

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" $\overrightarrow{F}$during given interval t is$P_{avg}=8.00\mathrm{W}$

(c) determining the angle between vectors   and  is,

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,

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

It is known that,

${\overrightarrow{d}}_i.\overrightarrow{d_f}=d_i{d}_f\cos \left(\varphi \right)$

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

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

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

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.