# 7.1 Work  (Page 7/11)

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## Work done by a spring force

A perfectly elastic spring requires 0.54 J of work to stretch 6 cm from its equilibrium position, as in [link] (b). (a) What is its spring constant k ? (b) How much work is required to stretch it an additional 6 cm?

## Strategy

Work “required” means work done against the spring force, which is the negative of the work in [link] , that is

$W=\frac{1}{2}k\left({x}_{B}^{2}-{x}_{A}^{2}\right).$

For part (a), ${x}_{A}=0$ and ${x}_{B}=6\text{cm}$ ; for part (b), ${x}_{B}=6\text{cm}$ and ${x}_{B}=12\text{cm}$ . In part (a), the work is given and you can solve for the spring constant; in part (b), you can use the value of k , from part (a), to solve for the work.

## Solution

1. $W=0.54\phantom{\rule{0.2em}{0ex}}\text{J}=\frac{1}{2}k\left[{\left(6\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}-0\right]$ , so $k=3\phantom{\rule{0.2em}{0ex}}\text{N/cm}\text{.}$
2. $W=\frac{1}{2}\left(3\phantom{\rule{0.2em}{0ex}}\text{N/cm}\right)\left[{\left(12\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}-{\left(6\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}\right]=1.62\phantom{\rule{0.2em}{0ex}}\text{J}.$

## Significance

Since the work done by a spring force is independent of the path, you only needed to calculate the difference in the quantity $½k{x}^{2}$ at the end points. Notice that the work required to stretch the spring from 0 to 12 cm is four times that required to stretch it from 0 to 6 cm, because that work depends on the square of the amount of stretch from equilibrium, $½k{x}^{2}$ . In this circumstance, the work to stretch the spring from 0 to 12 cm is also equal to the work for a composite path from 0 to 6 cm followed by an additional stretch from 6 cm to 12 cm. Therefore, $4W\left(0\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}6\phantom{\rule{0.2em}{0ex}}\text{cm}\right)=W\left(0\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}6\phantom{\rule{0.2em}{0ex}}\text{cm}\right)+W\left(6\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}12\phantom{\rule{0.2em}{0ex}}\text{cm}\right)$ , or $W\left(6\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}12\phantom{\rule{0.2em}{0ex}}\text{cm}\right)=3W\left(0\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}6\phantom{\rule{0.2em}{0ex}}\text{cm}\right)$ , as we found above.

Check Your Understanding The spring in [link] is compressed 6 cm from its equilibrium length. (a) Does the spring force do positive or negative work and (b) what is the magnitude?

a. The spring force is the opposite direction to a compression (as it is for an extension), so the work it does is negative. b. The work done depends on the square of the displacement, which is the same for $x=±6\phantom{\rule{0.2em}{0ex}}\text{cm}$ , so the magnitude is 0.54 J.

## Summary

• The infinitesimal increment of work done by a force, acting over an infinitesimal displacement, is the dot product of the force and the displacement.
• The work done by a force, acting over a finite path, is the integral of the infinitesimal increments of work done along the path.
• The work done against a force is the negative of the work done by the force.
• The work done by a normal or frictional contact force must be determined in each particular case.
• The work done by the force of gravity, on an object near the surface of Earth, depends only on the weight of the object and the difference in height through which it moved.
• The work done by a spring force, acting from an initial position to a final position, depends only on the spring constant and the squares of those positions.

## Conceptual questions

Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy transferred or changed in form in your example? If so, explain how this is accomplished without doing work.

When you push on the wall, this “feels” like work; however, there is no displacement so there is no physical work. Energy is consumed, but no energy is transferred.

Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work.

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