I wonder what the scale would read?

Question

A spring scale is attached to the ceiling and the other end is attached a stretched spring which in turn is attached to the floor. The spring scale reads 15.8 kg.
I wonder what the scale would read if:
(a) Some one decides to hang a mass of 10 kg off that scale
(b) Then comes back and attaches another 10 kg.

The spring constant is 295.7 N / m

...for both springs

1. So far Leonardo D leads the answer. He was the first and who also concisely answered the question. I wish Leonardo D expanded on concept of preloading.
2. Bekki B has frequently provided wonderful answers to the forum and who is justly leading the Physics category in quality and number of best answers. You touched me by saying “Simple spring problems confuse me for some reason I don't know.” They confuse me too ;-).
3. The Kat...Hmmm... What a fighter. Just for that valiant fight put up by you I’m tempted to press the ‘Best Answer’ button against you name. Good thinking!
The best way to look at this problem is by assume that you stretched a spring between say two nails driven into board. The force forcing either of the nails is F=-kx. One has to exceed this force by ..., yes you are correct, by an amount of F. So if you exceed F by an amount of df the spring will be stretched F + df = -k(x +dx). I will confess that it took me an evening, a seemingly very long evening..:-)

Answer 1

a) would still read 15.8 Kg because you have not exceeded the preload in the spring

b) would read 20Kg, since suspended weight is high than the preload

Answer 2

Ok, that question is really annoying me now!

RE*EDITED VERSION*EDITED VERSION* (third answer...)
The two thumbs pointing in different directions actually made me realize something.
Maybe I'm still totally off track, but I have nothing to lose!

When the first 10 kg was hung, the bottom spring "lost" 10kg
to the spring scale, so (a) the spring scale still reads 15.8.

When the segond mass of 10kg is added, the bottom spring
gives its last 5.8 kg , it has regained its original shape and prevents the spring scale from stretching any further.
It stops there, there should be a 20kg reading if the bottom spring was absent.

The scale has hit rock bottom! (b) 20-5.8=14.2kg
Thank you for posting such an annoying question!!

Answer 3

(a) It depends on the properties of the spring in the spring scale and the properties of the spring below. If the spring on the scale is super stiff, adding more mass won't change position much, so it won't change the pull of the spring below--you just add 10kg and 10 more kg.

If the spring on the scale is looser, adding weight to it will pull it down and relax the pull of the spring below, so it doesn't go down as much and you read a lighter weight.

So you need to know more.

Edit--if the springs are equal, then the total spring constant is doubled . You effectively have two springs in parallel, so they add. So if I add a given weight, I get half the displacement. So adding 10kg increases the reading by 5kg. Adding 10kg more increases it by 5kg more after that. That's my final answer :) Simple spring problems confuse me for some reason I don't know.