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Unread 16-05-2005, 02:30
sanddrag sanddrag is offline
On to my 16th year in FRC
FRC #0696 (Circuit Breakers)
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Join Date: Jul 2002
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Re: Springs and Work

Quote:
Originally Posted by eugenebrooks
At this point, you subtract the work for a stretch of (40-30) cm
from the work for (35-30) cm and you get the difference.
Okay, this is helpful. I never thought of this approach. I assume you meant 42 instead of 40. Anyway, how do I find the work for 30-35?

I assumed I would need to solve this problem by finding the spring constant k (you called it a, any letter will do)

When I solve for this I get 277.7777 N/m for the spring constant

when I try to find the work from 30-35, I assume we can set 30 = to 0 because it is the natural length of the spring. Then 35 would actually be 5 cm of change in length. So, if we put .05 meters into Work = (1/2)(kx^2) I get .347 N-m or Joules of Work. If I subtract this from 2 Joules of Work for an extend from 30-42 (or 0-12 depending on how you look at it) I don't get 1.04, I get 1.65

Any ideas?
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