Solving 2 Blocks on a Spring: Ma = Kx

[eri139]

Homework Statement

A 5 kilogram block is fastened to a vertical spring that has a spring constant of 1300 newtons per meter. A 3 kilogram block rests motionless on top of the 5 kilogram block.

a. When the blocks are at rest, how much is the spring compressed from its original length?

Relevant Equations

F = ma, F = kx

I honestly have no idea why I'm getting this question wrong, because it seems fairly straightforward.

I thought that treating the two blocks as one object would work, so 5 + 3 = 8. With Fg = Fs, ma = kx. Then, 8 x 9.8/1300 = x. x would be 0.06. That's been marked as an incorrect answer, though.

Thank you for any help!

edit: I got it! the issue was with preciseness, I didn't enter enough digits

 

[I have a Dream]

well for (a), please forgive me if I am wrong, but I am visualizing the problem as if you placed the 3kg block on top of the 5kg spring system. Therefore, the original length was when the 5kg spring system had no extra weight and the compressed length was when you placed the 3kg block on top i.e. m = 3 not m = 5 + 3

 

[eri139]

well for (a), please forgive me if I am wrong, but I am visualizing the problem as if you placed the 3kg block on top of the 5kg spring system. Therefore, the original length was when the 5kg spring system had no extra weight and the compressed length was when you placed the 3kg block on top i.e. m = 3 not m = 5 + 3

I thought you had it for a moment! That would make a lot of sense. But unfortunately I just tried it and it's still incorrect :(

 

[I have a Dream]

I thought you had it for a moment! That would make a lot of sense. But unfortunately I just tried it and it's still incorrect :(

So sorry! As my last comment I feel as if working with the elastic potential energy equation
For springs: PE = (1/2)kx^(2) is worth a shot. Since PE Is also equal to mgh or mgx in this case. Keep on working hard! You got this!