1) A force of S N to is required to compress an industrial spring by 0.000] meters, how much force is required to compress this spring by 0.0003 meters? 2. You have a certain set-up of a vertical spring, and when hung ?'eely 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

The first question uses hooks law which states that F=kx .

Thus we see that force is directly related to displacement and if we triple the displacement the force could be tripled as well. So it requires 15N.

you know that the frequency is one over the period (1/T).

and the period of a spring is given by:

T=2 pi sqrt((m)/(k))->

F=(1)/(2 pi)sqrt((k)/(m))->

k=(2 pi F)^(2)m=631.65m

now when the object is just hanging at rest there are two opposite forces on it that have to be equal to give a net force of zero that is the restoring force and the wieght so:

kx=mg

631.65mx=mg

631.65x=g

x=(9.8)/(631.65)=0.0155m

Answer-2

It will be less - and depending on your terminology it will be around 5cm.

Is the Equilibrium position the extension (compared to horizontal) of the spring just under it's own mass / if so that extension will be around 5cm on the Moon.

The extension experienced by a spring is directly proportional to the forces acting on the spring (within the normal range of the spring).

Since at the surface of the Moon the gravitational attraction g is about 1/6 of the value of g on Earth; that means for any given mass the extension on the Moon will be 1/6 of the extension experienced on Earth with the same Mass.

Remember even with no mass being hung on the spring, the spring is still affected (and extended) by it's own mass when hung vertically.