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>0when the spring is stretched andx<0when it is compressed. How much work must be done (a) to stretch this spring by 0.050 mfrom its unstretched length? (b) To compress this spring by 0.050 mfrom 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.)

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.

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

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 .

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.

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.