Q15E - University Physics with Modern Physics

Question

On a farm, you are pushing on a stubborn pig with a constant horizontal force with a magnitude 30.0 N and a direction 37.0⁰ counterclockwise from the +x-axis. How much work does this force do during a displacement of the pig that is (a) $\mathbf{s}\mathbf{=}\left(5.00\right)\hat{\mathbf{i}}$ ; (b)  $\mathbf{s}\mathbf{=}\left(‐6.00\right)\hat{\mathbf{j}}$; (c) $\mathbf{s}\mathbf{=}\mathbf{‐}\left(2.00\right)\hat{\mathbf{i}}\mathbf{+}\left(5.00\right)\hat{\mathbf{j}}$ .

Step-by-Step Solution

Answer

(a) 119.8 J

(b) -108.3 J

(c) 24.28 J

Step 1: Identification of the given data

The given data is listed below as-

• The direction is, 37⁰ counterclockwise from the x-axis.
• The constant horizontal Force applied is, $F=30.0 N$

Step 2: Significance of the work done

Work done on a particle by a constant force $\overrightarrow{\mathbf{F}}$  during a linear displacement $\overrightarrow{\mathbf{s}}$ 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 by the force during a displacement of the pig that is s → = ( 5   m ) i ^ (a)

The force vector of constant horizontal force in a direction that is 37⁰ counterclockwise from the x-axis will be

$$\begin{gathered} F=\left(30 \cos 37° N\right)\hat{i}+\left(30 \sin 37° N\right)\hat{j} \\ F=\left(23.96 N\right)\hat{i}+\left(18.05 N\right)\hat{j} \end{gathered}$$ 

Now,   $W=Fs \cos \varphi$

Or, $W=\overrightarrow{F}·\overrightarrow{s}$ 

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)

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

Now, Work done by the force during a displacement of $\overrightarrow{s}=\left(5 m\right)\hat{i}$ ,

For $s_x=5 m, s_y=0 m, F_x=23.96 N$ and  $F_y=18.05 N$

  $$\begin{gathered} W=\left(5 m\times 23 N\right)+\left(0 m\times 18.05 N\right) \\ =\left(119.8 Nm\right) \\ =119.8 J \end{gathered}$$

Thus, work done by the force during a displacement of the pig that is $\overrightarrow{s}=\left(5 m\right)\hat{i}$ is 119.8 J.

Step 4: Determination of work done by the force during a displacement of the pig that is s → = ( ‐ 6   m )   j ^ (b)

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

Now, Work done by the force during a displacement of $\overrightarrow{s}=\left(-6 m\right) \hat{j}$ ,

For $s_x=0 m, s_y=-6 m, F_x=23.96 N$ and  $F_y=18.05 N$

  

$$\begin{gathered} W=\left(0 m\times 23 N\right)+\left(-6 m\times 18.05 N\right) \\ =\left(-108.3 Nm\right) \\ =-108.3 J \end{gathered}$$

Thus, work done by the force during a displacement of the pig that is $\overrightarrow{s}=\left(-6 m\right) \hat{j}$ is

 -108.3 J.

Step 5: Determination of work done by the force during a displacement of the pig that is s → = ‐ ( 2   m )   i ^ + ( 4   m )   j ^ (c)

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

Now, Work done by the force during a displacement of $\overrightarrow{s}=-\left(2 m\right)\hat{i}+\left(4 m\right) \hat{j}$  ,

For  $s_x=-2 m, s_y=4 m , F_x=23.96 N$ and  $F_y=18.05 N$

$$\begin{gathered} W=\left(-2 m\times 23 N\right)+\left(4 m\times 18.05 N\right) \\ =\left(24.28 Nm\right) \\ =24.28 J \end{gathered}$$ 

 

Thus, work done by the force during a displacement of the pig that is $\overrightarrow{s}=-\left(-2 m\right) \hat{i}+\left(4 m\right) \hat{j}$  is 24.28 J.