Q62P - University Physics with Modern Physics

Question

 A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x=0 to x=6.9 m as you apply a force with x-component ${\mathbf{F}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{[}\mathbf{20}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{/}\mathbf{m}\mathbf{\left)}\mathbf{x}\mathbf{\right]}$. How much work does the force you apply do on the cow during this displacement?

Step-by-Step Solution

Answer

The work done applied by the force on cow is 209.4 J

Step 1: Identification of given data

The given data can be listed below,

a) The x component of force is, ${\mathrm{F}}_{\mathrm{x}}=-[20\hspace{0.33em}\mathrm{N}+(3\hspace{0.33em}\mathrm{N}/\mathrm{m}\left)\mathrm{x}\right]$ 

b) The cow moves from $\mathrm{x}=0\hspace{0.33em}\mathrm{to}\hspace{0.33em}\mathrm{x}=6.9\hspace{0.33em}\mathrm{m}$

Step 2: Concept/Significance of work done

If on the application of force on an object, it gets displaced from its initial position, the work is said to be done.

Step 3: Determination of the work applied by force on cow.

The work done by force on cow is given by,

$\mathrm{W}={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}{\mathrm{F}}_{\mathrm{x}}\mathrm{dx}$ 

Substitute all the values in the above,

$$\begin{gathered} \mathrm{W}={\int }_0^{6.9 \mathrm{m}}\left[20 \mathrm{N}+\left(3 \mathrm{N}/\mathrm{m}\right)\mathrm{x}\right]\mathrm{dx} \\ =-{{\left(\left(20\mathrm{N}\right)\mathrm{x}\right)}_0}^{6.9\mathrm{m}}-\left(3.0 \mathrm{N}/\mathrm{m}\right)\left(\frac{1}{2}\right){\left(6.9\mathrm{m}\right)}^2 \\ =-138 \mathrm{N}.\mathrm{m}-71.4 \mathrm{N} \\ =-209.4 \mathrm{J} \end{gathered}$$ 

Thus, the work done applied by the force on cow is $209.4 \mathrm{J}$