Q70P - University Physics with Modern Physics

Question

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount x, a force along the x-axis with x-component ${\mathbf{F}}_{\mathbf{x}}\mathbf{=}\mathbf{kx}\mathbf{-}{\mathbf{bx}}^{\mathbf{2}}\mathbf{+}{\mathbf{cx}}^{\mathbf{3}}$ must be applied to the free end. Here $\mathbf{k}\mathbf{=}\mathbf{100}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{/}\mathbf{m}\mathbf{,}\mathbf{\hspace{0.33em}}\mathbf{b}\mathbf{=}\mathbf{700}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$and $\mathbf{c}\mathbf{=}\mathbf{12}\mathbf{,}\mathbf{000}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$. Note that x>0 when the spring is stretched and x<0 when it is compressed. How much work must be done (a) to stretch this spring by 0.050 m from its unstretched length? (b) To compress this spring by 0.050 m from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of ${\mathit{F}}_{\mathbf{x}}$ on x. (Many real springs behave qualitatively in the same way.)

Step-by-Step Solution

Answer

(a) The work done in stretching the spring from zero to 0.50 m is 0.115 J . 

(b) The work done in compressing the spring from zero to 0.050 m is 0.173 J

(c) More work is required to compress the spring than stretching for the same distance.

Step 1: Definition of Hooke&rsquo;s law

Hooke’s law states that the restoring force is always directly proportional to the amount of stretch/deformation in the spring, such that if deformation in the spring is x, then restoring force in the spring will be,

 

$\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{kx}$ 

 

where k is a positive constant.

Step 2: Given data

$$\begin{gathered} {\mathrm{F}}_{\mathrm{x}}=\mathrm{kx}-{\mathrm{bx}}^2+{\mathrm{cx}}^3 \\ \mathrm{k}=100\hspace{0.33em}\mathrm{N}/\mathrm{m} \\ \mathrm{b}=700\hspace{0.33em}\mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{c}=12,000\hspace{0.33em}\mathrm{N}/{\mathrm{m}}^3 \end{gathered}$$

Step 3: (a) Find the unstretched length

Use the concept of work done by a variable force that uses the method of integration.

   

The work done by the variable force is given as follows:

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

 

Substitute 0 for $x_1$, 0.050 m  for $x_2$ , and $kx-b{x}^2+c{x}^3$ for $F_x$ .

$$\begin{gathered} {\mathrm{W}}_{\mathrm{stretching}}={\int }_0^{0.050}\left(\mathrm{kx}-{\mathrm{bx}}^2+{\mathrm{cx}}^3\right)\mathrm{dx} \\ ={\left[\mathrm{k}\left(\frac{{\mathrm{x}}^2}{2}\right)-\mathrm{b}\left(\frac{{\mathrm{x}}^3}{3}\right)+\mathrm{c}\left(\frac{{\mathrm{x}}^4}{4}\right)\right]}_0^{0.050 \mathrm{m}} \\ =\mathrm{k}\left(\frac{{\left(0.050 \mathrm{m}\right)}^2}{2}-0\right)-\mathrm{b}\left(\frac{\left(0.050 \mathrm{m}\right)}{3}-0\right)+\mathrm{c}\left(\frac{\left(0.050 \mathrm{m}\right)}{4}-0\right) \\ =\left(100\mathrm{N}/\mathrm{m}\right)\frac{{\left(0.050 \mathrm{m}\right)}^2}{2}-\left(700 \mathrm{N}/{\mathrm{m}}^2\right)\frac{\left(0.050 \mathrm{m}\right)}{3}+\left(12,000\mathrm{N}/{\mathrm{m}}^3\right)\frac{{\left(0.050\right)}^4}{4} \\ =0.115 \mathrm{JA} \end{gathered}$$ 

  

 

Therefore, the work done in stretching the spring from zero to  

Is 0.115 J . 

Step 4: (b) Find the unstretched length

The work done by the variable force is given as follows:

  

 

Substitute 0 m for $x_1$, $-0.050 \mathrm{m}$ for $x_2$ , and $\mathrm{kx}-{\mathrm{bx}}^2+{\mathrm{cx}}^3$ for ${\mathrm{F}}_{\mathrm{x}}$ .

$$\begin{gathered} {\mathrm{w}}_{\mathrm{compress} }={\int }_0^{-0.050\mathrm{m}} \left(\mathrm{kx}-{\mathrm{bx}}^2+{\mathrm{cx}}^3\right)\mathrm{dx} \\ ={\left[\mathrm{k}\left(\frac{{\mathrm{x}}^2}{2}\right)-\mathrm{b}\left(\frac{{\mathrm{x}}^3}{3}\right)+\mathrm{c}\left(\frac{{\mathrm{x}}^4}{4}\right)\right]}_0^{-0.050\mathrm{m}} \\ =\mathrm{k}\left(\frac{(-0.050\mathrm{m}{)}^2}{2}-0\right)-\mathrm{b}\left(\frac{(-0.050\mathrm{m})}{3}-0\right)+\mathrm{c}\left(\frac{(-0.050\mathrm{m}{)}^4}{4}-0\right) \\ =(100\mathrm{N}/\mathrm{m})\frac{(0.050\mathrm{m}{)}^2}{2}+\left(700\mathrm{N}/{\mathrm{m}}^2\right)\frac{(0.050\mathrm{m}{)}^3}{3}+\left(12,000\mathrm{N}/{\mathrm{m}}^3\right)\frac{(0.050\mathrm{m}{)}^4}{4} \\ =0.173\mathrm{J} \end{gathered}$$ 

  

 

Therefore, the work done in compressing the spring from zero to 0.050 m   

is 0.173 J. 

Step 5: (c) Find is it easier to stretch or compress this spring

More work is required to compress the spring than stretching for the same distance. 

This is because of the reason that the work is the function of polynomials of x and one term $-{\mathrm{bx}}^2$ is negative which makes compression tougher.