Question

Two identical, side-by-side springs with spring constant 280 N/m support a 1.50kg hanging box. By how much is each spring stretched?

Answer 1

Approximately
`0.0263\; {\rm m}` (assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.)

Step-by-step explanation:

The weight of this
`m = 1.50\; {\rm kg}` box is:

`\begin{aligned}(\text{weight}) &= m\, g \\ &= (1.50\; {\rm kg})\, (9.81\; {\rm N\cdot kg^(-1)}) \\ &\approx 14.715\; {\rm N}\end{aligned}`.

The tensions in the two springs, combined, is equal to the weight of this box.

To determine how the weight is split between the two springs, consider Hooke's Law. By Hooke's Law, the tension in the spring is proportional to the distance (displacement) by which the spring is stretched:

`(\text{tension}) = (\text{spring constant})\, (\text{displacement})`.

The spring constant of the two springs in this questions are equal. For the springs to hold the box side-by-side, their displacements also need to be equal. Hence, the tension in the two springs would be the same.

Therefore, the weight of the box would be evenly split between the two springs. The tension in each spring would be:

`\begin{aligned}\frac{(\text{weight})}{2} &= \frac{14.715\; {\rm N}}{2} \approx 7.3575\; {\rm N}\end{aligned}`.

To find the displacement of the spring, divide the tension in each spring by the spring constant:

`\begin{aligned}(\text{displacement}) &= \frac{(\text{tension})}{(\text{spring constant})} \\ &\approx \frac{7.3575\; {\rm N}}{280\: {\rm N\cdot m^(-1)}} \\ &\approx 0.0263\; {\rm m}\end{aligned}`.