Question

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20cm long?

Answer 1

Approximately
`25\; {\rm N}` (assuming that this spring is ideal.)

Step-by-step explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

• When the
  `6.0\; {\rm cm}`-spring in this question is stretched to
  `10\; {\rm cm}`, the displacement is
  `x = (10\; {\rm cm} - 6.0\; {\rm cm})`.
• Likewise, if this spring is stretched to
  `20\; {\rm cm}`, the displacement would be
  `(20\; {\rm cm} - 6\; {\rm cm})`.

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of
`F_{\text{a}}` displaces this spring by
`x_{\text{a}}`, while a force of
`F_{\text{b}}` displaces this spring by
`x_{\text{b}}`, then:

`\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}`.

In this question, it is given that a force of
`F_{\text{a}} = 7.0 \; {\rm N}` would stretch this spring by
`x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})`. Thus, the force
`F_{\text{b}}` required to stretch this spring by
`x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})` would satisfy:

`\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}`.

Rearrange and solve for
`F_{\text{b}}`:

`\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}`.