
#1
Jun307, 02:39 PM

P: 2

1. The problem statement, all variables and given/known data
A mass m hangs from a spring with stiffness constant k. The spring is cut in half and the same mass hung from it. WIll the new arrangement have a higher or a lower stiffness constant than the original spring? 2. Relevant equations F= kx 3. The attempt at a solution I think that the spring will have the same spring stiffness. I dont feel that the spring constant is something that can be changed. If it was, then it shouldnt be called a constant. 



#2
Jun307, 02:48 PM

P: 1,705





#3
Jun307, 02:52 PM

P: 398

No, it is the spring constant. It does not change assuming it is the same spring. It will just half the amount of compression that the spring can take before you go past the limit.
On a side note, I think my arrow keys are disable in the forum and when I hit the apostrophe key I get a quick search link. Is anyone else having the same problem? 



#4
Jun307, 04:50 PM

P: 1,705

spring constant 



#5
Jun307, 05:04 PM

P: 398





#6
Jun307, 05:17 PM

P: 1,705

tempering i meant the tempering of the spring is affected 



#7
Jun307, 05:22 PM

P: 398

If it isn't a car spring I highly doubt that you would use a acetylene torch to cut it. Say a mechanical pencil spring




#8
Jun307, 05:24 PM

P: 1,705





#9
Jun307, 05:26 PM

P: 398





#10
Jun307, 05:31 PM

P: 1,705





#11
Jun307, 07:34 PM

P: 2

thanks everyone for the help




#12
Jun307, 07:35 PM

Mentor
P: 40,883

Another way to look at it: Hang a weight from the original spring and it stretches the spring a distance X. Now imagine that the spring is really two springs attached together (each one half the size). How much does that weight stretch each halfspring? 



#13
Jun307, 07:38 PM

P: 1,705




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