Question

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.

Short answer

a. Work required to stretch the string from x = 0.500m to x = 1.00 m is${\mathrm{W}}_{\mathrm{spring}}=31.0\mathrm{J}$

b. The speed of the particle when stretched from the x = 0.500 m is 5.35 m/s.

c. The spring force is conservative.

Step-by-step solution

Given

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

Understand the concept

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:

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

(a) Calculate the work required to stretch the spring from x = 0.500m to x = 1.00m.

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,

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

(b) Calculate the speed at the instant the stretch in the spring is x = 0.500m  

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

${\mathrm{K}}_{\mathrm{i}}=0$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.

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

(c) Explain if the force exerted by the spring is conservative or nonconservative

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.