Question

A 5 Kg mass is suspended from a spring. The spring is stretched 8 cm from equilibrium. What is the spring constant? Use g=9.8 m/s^2 to calculate the force of the load

Answer 1

The spring constant is:

                                             $\Large\displaystyle\text{ \begin{aligned}k &= 612.5\ \dfrac{\text{N}}{\text{m}}\end{aligned} }$

To calculate the spring constant we must remember the law for it, the Hooke's Law:

                                            $\Large\displaystyle\begin{aligned}\vec{F} = -k\Delta \vec{x} \end{aligned}$

Where k is the spring constant [N/m].

So if the mass is suspended it means that its weight is equal to the elastic force (values), then we can write:

                                           $\Large\displaystyle\begin{aligned}\vec{W} = k\Delta \vec{x} \end{aligned}$

Therefore:

                                          $\Large\displaystyle\begin{aligned}mg &= k\Delta x \\ \\k &= \dfrac{mg}{\Delta x} \\ \\\end{aligned}$

Now we just have to put the values and calculate:

                                              $\Large\displaystyle\text{ \begin{aligned}k &= \dfrac{mg}{\Delta x} \\ \\k &= \dfrac{5\cdot 9.8}{0.08} \\ \\k &= 612.5\ \dfrac{\text{N}}{\text{m}}\end{aligned} }$

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