A certain spring is found not to conform to Hooke’s law. The force (in newtons) it exerts when stretched a distance x(in meters) is found to have magnitude 52.8x +38.4${\mathit{x}}^{\mathbf{2}}$in the direction opposing the stretch. (a) Compute the work required to stretch the spring from x =0.500mto x = 1.00m. (b) With one end of the spring fixed, a particle of mass 2.17 kgis attached to the other end of the spring when it is stretched by an amount x = 1.00m. If the particle is then released from rest, what is its speed at the instant the stretch in the spring is x = 0.500m? (c) Is the force exerted by the spring conservative or nonconservative? Explain.

Force applied by the spring when stretched for a distance x is F =$52.8x+38.4x^2$

The mass of the particle is m = 2.17 kg

Stretching due to the mass is x = 1.00 m

By finding the integral of force and displacement we can find work done by the spring.

Using the value of work done we can find the velocity of the particle.

Using$\mathbf{W}\mathbf{=}\mathbf{∆}\mathbf{K}\mathbf{+}\mathbf{∆}\mathbf{U}$we can find that the spring force is conservative or non -conservative.

Formula:

$$\begin{gathered} \mathbf{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}\mathbf{Fdx} \\ \mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}} \\ \mathbf{U}\mathbf{=}\mathbf{mgh} \end{gathered}$$

The work done can be estimated using the formula,

$\mathbf{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}\mathbf{Fdx}$

Where F is the force and work done is estimated between the points x₁ and x₂. Therefore,

$$\begin{gathered} \mathrm{W}=\left[{\int }_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{fdx}\right]\mathrm{J} \\ =\left[{\int }_{0.5}^1(52.8\mathrm{x}+38.4{\mathrm{x}}^2)\mathrm{dx}\right]\mathrm{J} \\ =\left[{\int }_{0.5}^{1}52.8\mathrm{xdx}\right]\mathrm{J}+\left[{\int }_{0.5}^{1}38.4{\mathrm{x}}^2\mathrm{dx}\right]\mathrm{J} \\ =\left({\left[\frac{52.8}{2}{\mathrm{x}}^2\right]}_{0.5}^1\right)\mathrm{J}+\left({\left[\frac{38.4}{3}{\mathrm{x}}^2\right]}_{0.5}^1\right)\mathrm{J} \\ =(26.4\left[1^2-0.5^2\right])\mathrm{J}+(12.8\left[1^3-0.5^3\right])\mathrm{J} \\ =31.0\mathrm{J} \end{gathered}$$

$$\begin{gathered} \mathrm{W}=∆\mathrm{K} \\ ={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} \end{gathered}$$

${\mathrm{K}}_{\mathrm{i}}=0$since theparticle is released from rest, therefore, the above expression can be written as,

$$\begin{gathered} \mathrm{W}={\mathrm{k}}_{\mathrm{f}} \\ =\frac{1}{2}{\mathrm{mv}}^2 \end{gathered}$$

Rearranging the equation for velocity.

$$\begin{gathered} \mathrm{v}=\sqrt{\left(\frac{2\mathrm{W}}{\mathrm{m}}\right)} \\ =\sqrt{\left(\frac{2\times 31.0\mathrm{J}}{2.17\mathrm{kg}}\right)} \\ =5.35\mathrm{m}/\mathrm{s} \end{gathered}$$

Force is conservative since work done by the force is independent of the path it followed, it depends on the initial and final position of the particle.