Question

One end of a spring whose spring constant is 20N/mis attached to the wall, and you pull on the other end, stretching it from its equilibrium length of 0.2mto a length of 0.3m. Estimate the work done by dividing the stretching process into two stages and using the average force you exert to calculate work done during each stage.

Short answer

The work done is 0.5 J .

Step-by-step solution

Identification of the given data

The given data is listed as follows,

• The spring constant of the one end of a spring is,$k=20\mathrm{N}/\mathrm{m}$
• The initial equilibrium length of the spring is,$x_1=0.2\mathrm{m}$
• The final equilibrium length of the spring is,$x_3=0.3\mathrm{m}$

Significance of the work done

The work done is equal to the force exerted on an object and the distance the object has to move. The expression for the work done is as follows,

W=F . d

Here, is the force and is the distance.

The equation of the work done gives the total work done.

Determination of the work done in the first stage

$x$As the stretching process is being divided, then the middle equilibrium length of the spring will be 0.25m as it is the midpoint between the initial and the final equilibrium length.

The equation of the work done in the first stage is expressed as:

$W_1=\underset{x_1}{\overset{x_2}{\int }}F.dr$

Here, $x_2$is the equilibrium length at the midpoint, $x_1$is the initial equilibrium length of the spring, $F$is the force exerted and $dr$is the rate of change of the radial distance.

The above equation can also be expressed as:

role="math" localid="1657086679515" $W_1=\underset{x_1}{\overset{x_2}{\int }}Kx.dx$

Here,$k$ is the spring constant,$dx$ is the radial distance and is the change in the equilibrium length.

Substitute 20 N/m for $k$,0.2m for $x_1$and 0.25m for $x_2$in the above equation.

role="math" localid="1657087187799" $$\begin{gathered} W_1=20N/m\underset{0.2m}{\overset{0.25m}{\int }}x.dx \\ =20\mathrm{N}/\mathrm{m}\times \frac{1}{2}{\left[{\mathrm{x}}^2\right]}_{0.2\mathrm{m}}^{0.25\mathrm{m}} \\ =10\mathrm{N}/\mathrm{m}\times \frac{1}{2}\left[0.0625{\mathrm{m}}^2-0.04{\mathrm{m}}^2\right] \\ =10\mathrm{N}/\mathrm{m}\times 0.0225{\mathrm{m}}^2 \\ =0.0225\mathrm{N}.\mathrm{m} \\ =0.225\mathrm{N}.\mathrm{m}\times \left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \\ =0.225\mathrm{J} \end{gathered}$$

Determination of the work done in the second stage

The equation of the work done in the second stage is expressed as:

$W_1=\underset{x_1}{\overset{x_2}{\int }}F.dr$

Here,$x_2$ is the equilibrium length at the midpoint, and$x_3$ is the final equilibrium length of the spring.

The above equation can also be expressed as:

$W_2=\underset{x_2}{\overset{x_3}{\int }}Kx.dx$

Substitute 20N/m for $k$,0.3m for $x_3$and 0.25 m for $x_2$in the above equation.

$$\begin{gathered} W_2=20N/m\underset{0.25m}{\overset{0.3m}{\int }}x.dx \\ =20\mathrm{N}/\mathrm{m}\times \frac{1}{2}{\left[{\mathrm{x}}^2\right]}_{0.25\mathrm{m}}^{0.3\mathrm{m}} \\ =10\mathrm{N}/\mathrm{m}\times \left[0.09{\mathrm{m}}^2-0.0625{\mathrm{m}}^2\right] \\ =10\mathrm{N}/\mathrm{m}\times 0.0275{\mathrm{m}}^2 \\ =0.275\mathrm{N}.\mathrm{m} \\ =0.275\mathrm{N}.\mathrm{m}\times \left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \\ =0.275\mathrm{J} \end{gathered}$$

Determination of the total work done

The equation of the total work done is expressed as:

$W=W_1+W_2$

Substitute 0.275J for$W_2$ and 0.225 J for$W_1$ in the above equation.

$$\begin{gathered} W=0.225\mathrm{J}+0.275\mathrm{J} \\ =0.5\mathrm{J} \end{gathered}$$

Thus, the work done is 0.5J .