Answer to Question #296864 in Mechanics | Relativity for Sk Biswas

Question

You have a certain set-up of a vertical spring, and when hung freely the equilibrium position

reading is 30cm on earth. If you take the same set-up on the moon surface, would the equilibrium

position be more than, less than or equal to 30cm? Explain.

Answer 1

Explanations & Calculations

• At the equilibrium, it is possible to write $\small F=kx\implies mg=kx$ as the force generated on the spring is equal to the magnitude of the weight of the ganging mass.
• Then the extension is

$\qquad\qquad \begin{aligned} \small x&=\small \frac{mg}{k}\\ \small x&=\small \Big(\frac{m}{k}\Big)g \end{aligned}$

• It could be seen that the extension depends only on the gravitational acceleration of the area.

• On the moon, however, the gravitational acceleration is about $\small \frac{1}{6}$ of that on earth meaning less than that on earth.
• Therefore, the extension on the moon is lower compared to that on earth. (lower than cm)