Question

Find the spring constant for a spring that stretches 1.4cm when a 638 g weight is attached to it. How much work is done as the spring extends?

Answer 1

Spring constant:
`\rm 447\; N \cdot m^(-1)`.

Work done:
`\rm 0.0438\; J`.

Step-by-step explanation:

Convert all values to SI units.

• Length change:
  `\rm 1.4 \; cm = 0.014\; m`;
• Mass of the weight:
  `\rm 638\; g = 0.638\; kg`.

Assume that the spring-mass system is vertical and is placed on the surface of the earth. The gravitational acceleration constant will be equal to
`\rm 9.81\; N\cdot kg^(-1)`.

Gravitational pull on the weight:

`W = m\cdot g = \rm 0.638\; kg * 9.81\; N\cdot kg^(-1) = 6.25878\; N`.

That's also the size of the force on the spring.
`F = \rm 6.25878\; N`.

The spring constant is the size of the force required to deshape the spring (by stretching, in this case) by unit length.

`\displaystyle k_{\text{spring}} = (F)/(\Delta x) = \rm (6.25878\; N)/(0.014\; m) = 447.056\; N\cdot m^(-1)`.

Assume that there's no energy loss in this process. The work done on the spring is the same as the elastic potential energy that it gains:

`\begin{aligned} &\text{EPE} \\=& (1)/(2)k\cdot x^(2) \\=&\rm (1)/(2) * 447.056\; N\cdot m^(-1) * (0.014\; m)^(2)\\ =& \rm 0.0438\; J\end{aligned}`.