Question

6) A spring with a spring constant k = 800 N/m has been compressed, and 196 J of

potential energy is stored. What distance from equilibrium has the spring been
compressed?

Answer 1

`0.7\; {\rm m}`.

Step-by-step explanation:

If spring with spring constant
`k` is compressed by
`x` relative to the equilibrium, the elastic potential energy
`E` stored in that spring will be
`E = (1/2)\, k\, x^(2)`.

For the spring in this question, it is given that
`E = 196\; {\rm J}` whereas
`k = 800\; {\rm N \cdot m^(-1)}`. The displacement
`x` needs to be found.

Note that
`\begin{aligned}E &= 196\; {\rm J} = 196\; {\rm N \cdot m} \end{aligned}` (one joule is the work done when a force of
`1\; {\rm N}` is applied over a distance of
`1\; {\rm m}`.)

Rearrange the equation
`E = (1/2)\, k\, x^(2)` to find displacement
`x` in terms of potential energy
`E` and spring constant
`k`:

`\begin{aligned} x^(2) = (2\, E)/(k)\end{aligned}`.

`\begin{aligned} x &= \sqrt{(2\, E)/(k)} \\ &= \sqrt{\frac{2 * 196\; {\rm N \cdot m}}{800\; {\rm N\cdot m^(-1)}}} \\ &= \sqrt{0.49\; {\rm m^(2)}} \\ &= 0.7\; {\rm m} \end{aligned}`.

In other words, the spring has been compressed by
`x = 0.7\; {\rm m}` from the equilibrium.