Question #296864

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.


1
Expert's answer
2022-02-13T12:15:22-0500

Explanations & Calculations


  • At the equilibrium, it is possible to write F=kx    mg=kx\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

x=mgkx=(mk)g\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 16\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)

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS