You apply a constant force $\mathbf{F}\mathbf{=}\left(‐68.0N\right)\mathbf{}\hat{\mathbf{i}}\mathbf{+}\left(36.0N\right)\hat{\mathbf{j}}$ to the 380-kg car as the car travels 48.0 m in a direction that is 240⁰ counter clockwise from the +x-axis. How much work does the force you apply do on the car?

The given data is listed below as-

• The distance traveled by car is,$d=48m$
• The mass of the car is, $m=380kg$
• The constant Force applied is, $F=\left(-68N\right)\hat{i}+\left(36N\right)\hat{j}$

Work done on a particle during a linear displacement$\overrightarrow{\mathbf{s}}$by a constant force $\overrightarrow{\mathbf{F}}$is given by -

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{·}\overrightarrow{\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{Fs}\mathbf{}\mathbf{cos\varphi } \end{gathered}$$

If F and s are in the same direction then $\varphi =0$ and if it is in opposite direction then $\varphi =180°$.

The free-body diagram for the car is shown below:

The displacement vector when the car travels 48 m in a direction 240⁰counterclockwise from the x-axis will be

$$\begin{gathered} s=\left(48\cos 240°N\right)\hat{i}+\left(48\sin 240°N\right)\hat{j} \\ s=\left(-24m\right)\hat{i}+\left(-41.57m\right)\hat{j} \end{gathered}$$

Now, $W=Fs\cos \varphi$

Or,

Where,

$\overrightarrow{F}=F_x\hat{i}+F_y\hat{j}$………….(1)

And $\overrightarrow{s}=s_x\hat{i}+s_y\hat{j}$……..(2)

Now, $W=\overrightarrow{F}·\overrightarrow{s}$………..(3)

Therefore, substitute equation (1) and (2) in equation (3)

$$\begin{gathered} W=\overrightarrow{F}·\overrightarrow{s} \\ =\left(F_x\hat{i}+F_y\hat{j}\right)·\left(s_x\hat{i}+s_y\hat{j}\right) \\ =s_x{F}_x=s_y{F}_y \end{gathered}$$

For $s_x=-24m,s_y=-41.57m,F_x=-68N$ and $F_y=36N$

$$\begin{gathered} W=\left(-24m\times -68N\right)+\left(-41.57m\times 36N\right) \\ =\left(1632Nm\right)+\left(-1496.52Nm\right) \\ =135.48J \end{gathered}$$

Thus, the magnitude of work done by the given force on car is 135.48 J.