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

Answer:

Spring constant: $\rm 447\; N \cdot m^{-1}$.

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

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:

$W = m\cdot g = \rm 0.638\; kg \times 9.81\; N\cdot kg^{-1} = 6.25878\; N$.

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.

$\displaystyle k_{\text{spring}} = \frac{F}{\Delta x} = \rm \frac{6.25878\; N}{0.014\; m} = 447.056\; N\cdot m^{-1}$.

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:

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