91影视

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$\stackrel{\mathbf{鈫�}}{\mathbf{s}}$by a constant force $\stackrel{\mathbf{鈫�}}{\mathbf{F}}$is given by -

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{路}\stackrel{\mathbf{鈫�}}{\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{Fs}\mathbf{}\mathbf{肠辞蝉蠒} \end{gathered}$$

If F and s are in the same direction then $蠒=0$ and if it is in opposite direction then $蠒=180掳$.

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

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

Now, $W=Fs\cos 蠒$

Or,

Where,

$\stackrel{鈫�}{F}=F_x\hat{i}+F_y\hat{j}$鈥︹赌︹赌︹赌�.(1)

And $\stackrel{鈫�}{s}=s_x\hat{i}+s_y\hat{j}$鈥︹赌�..(2)

Now, $W=\stackrel{鈫�}{F}路\stackrel{鈫�}{s}$鈥︹赌︹赌�..(3)

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

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

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

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

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