Q14E - University Physics with Modern Physics

Question

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? 

Step-by-Step Solution

Answer

135.48 J

Step 1: Identification of the given data

The given data is listed below as-

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

Step 2: Significance of the work done

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 $\varphi =180°$.

Step 3: Determination of work done applied by given force on the car

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

           

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

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

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

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

 

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