Hooke's Law on the moon

[Deetle]

1. The problem statement, all variables and given/known data

How would F(g) and (delta) X change if the Spring experiment was done on the moon where the gravitational acceleration is six times smaller than on earth?

2. Relevant equations
F(g) = (delta) mg
F(s) = k(delta) X


3. The attempt at a solution

I would think that F(g) and (delta) X would be 6 times less on the moon because there would be less gravitational pull on the spring. So if F = 68600 g cm/sec^2 and X is 26.1 cm on earth, the F = 11433 g cm/sec^2and X = 4.35 cm on the moon.

Please help.I think I'm getting the concept confused. Thanks!

[t-money]

I don't think delta X would change, since that is a problem of conservation of energy assuming the system is in the horisontal position.

[arildno]

1. The problem statement, all variables and given/known data

How would F(g) and (delta) X change if the Spring experiment was done on the moon where the gravitational acceleration is six times smaller than on earth?

2. Relevant equations
F(g) = (delta) mg
F(s) = k(delta) X


3. The attempt at a solution

I would think that F(g) and (delta) X would be 6 times less on the moon because there would be less gravitational pull on the spring. So if F = 68600 g cm/sec^2 and X is 26.1 cm on earth, the F = 11433 g cm/sec^2and X = 4.35 cm on the moon.

Please help.I think I'm getting the concept confused. Thanks!
You are indeed right, for the hanging spring.

The only (problem-relevant) change in this lunar setup with respect to a tellar one is the change in the value of local g, the mass and the spring constant remaining the same.

Since lunar g is one sixth of tellar g, the lunar delta will be one sixt of the delta on Earth.

[HallsofIvy]

Yes, obviously if g on the moon is "6 times smaller than on earth" (1/6 the value) and the mass remains the same, then F(g) is 1/6 what it is on earth. (Actually it works the other way- because F(g) is 1/6 what it is on earth, g is 1/6.)
On earth, F= mg. Dividing both sides of that equation by 6, (F/6)= m(g/6).

But the "spring" question is not a clear. If the spring is lying on a flat surface, gravity plays no part. Are you assuming that a weight is hanging from the spring and gravitational force is stretching it, then the same thing happens. If F is the gravitational force on earth and F= kX, then dividing both sides by 6, (F/6)= k(X/6). Because, on the moon, the gravitational force if F/6, we also have the "stretch" equal to X/6.

[Deetle]

I believe that it is in a vertical position since the weight (mass) is hanging on the end of a spring and the spring is attached to the ceiling of a room.

[Deetle]

Thank you Halls of Ivy.

I'm still unclear about X. The formula on the moon is (F/6) = K (X/6).. Is this correct?

[Shooting Star]

I'm unclear about you problem. Use basic concepts.

When you hang a mass m on earth, and the extension is x, then force = mg = kx.

When you hang the same mass m on moon, and the extension is x2, then m(g/6) = kx2, which gives, x2 = x/6.