7.1 Work  (Page 7/11)

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

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

Significance

Since the work done by a spring force is independent of the path, you only needed to calculate the difference in the quantity $\mathrm{½}k{x}^2$ 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, $\mathrm{½}k{x}^2$ . 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, $4W(0\text{cm}\text{to}6\text{cm})=W(0\text{cm}\text{to}6\text{cm})+W(6\text{cm}\text{to}12\text{cm})$ , or $W(6\text{cm}\text{to}12\text{cm})=3W(0\text{cm}\text{to}6\text{cm})$ , as we found above.

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