Question

A 10kg block is placed onto a vertical spring with a spring constant of 250 N/m initially in its equilibrium position. How much does the spring compress and

how much potential energy is stored in the spring at that point?

Answer 1

The spring would compress by approximately
`0.39\; {\rm m}`.

The elastic potential energy stored in the spring would be approximately
`19\; {\rm J}`.

(Assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.)

Step-by-step explanation:

Assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`, the weight of this block (
`m = 10\; {\rm kg}`) would be:

`\begin{aligned} (\text{weight}) &= m\, g \\ &= (10\; {\rm kg})\, (9.81\; {\rm N\cdot kg^(-1)}) \\ &= 98.1\; {\rm N} \end{aligned}`.

The spring compresses under the weight of this block. Dividing the external force
`F` by the spring constant
`k` would give the displacement
`x` from equilibrium:

`\begin{aligned}(\text{displacement}) &= \frac{(\text{external force})}{(\text{spring constant})}\end{aligned}`.

`\begin{aligned}x &= (F)/(k) \\ &= \frac{98.1\; {\rm N}}{250\; {\rm N \cdot m^(-1)}} \approx 0.392\; {\rm m}\end{aligned}`.

When a spring of spring constant
`k` is compressed by a displacement of
`x`, the elastic potential energy
`(\text{EPE})` stored in this spring would be:

`\begin{aligned}(\text{EPE}) &= (1)/(2)\, (\text{spring constant})\, (\text{displacement})^(2)\end{aligned}`.

For the spring in this question:

`\begin{aligned}(\text{EPE}) &= (1)/(2)\, k\, x^(2) \\ &= (1)/(2)\, (250\; {\rm N\cdot m^(-1)})\, (0.392\; {\rm m})^(2) \\ &\approx 19\; {\rm J}\end{aligned}`.