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Work done on Spring

  • Thread starter TG3
  • Start date
  • #1
TG3
66
0

Homework Statement


Suppose 2 J is needed to stretch a spring from it's natural length of 35 cm to 42 cm. How much work is needed to stretch it from 35 cm to 40 cm?

Homework Equations


W=FD
F = KX

The Attempt at a Solution



2 = 12K
K= 1/6

Work = Integral from .35 to .4 of (1/6)X
(1/6)(x^2/2) from .35 to .4.
.013 - .010 = .033
Wrong.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
W=1/2 kx^2

so from 35 cm to 42 cm, the extension is 7 cm. The work done in this extension is 2J

find k.

Now when you get the value for k, find the extension from 35cm to 40cm and put it back into the formula W=1/2kx2
 
  • #3
137
4
Recognising that W is proportional to x^2 you could write the ratio of the known and unknown energies equal to the square of the ratio of the two extensions. The unknown spring constant vanishes in the division so you don't have to worry about calculating it.

2/W = (0.07)^2 / (0.05)^2
 

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