spring question

[swamprat]

Q: A spring has a spring constant of 100 N/cm. How much work is required to alter the spring from a compression of 10 cm to a compression of 50 cm?

My Answer:

F=-kx so F=-(100 N/cm)(40 cm) = 4,000 N

Then: W = Fd = (4,000 N)(.4 m) = 1600 J

The actual answer is 1200 J .. what did I mess up?

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[not respect]

the force is not constant like you put it in there, so you can't use those equations, if you use a little bit of calculus it should come out right... or just use the simpler equation.

 

[swamprat]

the force is not constant like you put it in there, so you can't use those equations, if you use a little bit of calculus it should come out right... or just use the simpler equation.

well you dont need calculus based physics for the mcat so what would the "simpler" equation be?

 

[BerkReviewTeach]

well you dont need calculus based physics for the mcat so what would the "simpler" equation be?

work = change in potential energy

0.5 x k (delta x)exp2 final - 0.5 x k (delta x)exp2 initial

make sure to convert to mks units where k = 10,000 N/m and distances are 0.5 m and 0.1 m

works out to be:

• 0.5 x 10000 N/m x .5exp2 - 0.5 x 10000 N/m x .1exp2 =
  
  0.5 x 10000 N/m x .25 - 0.5 x 10000 N/m x .01 =
  
  0.125 x 10000 - 0.005 x 10000 = 0.12 x 10000 = 1200 J

 

[Vihsadas]

This is a very subtle, but important point here, so please pay attention to BRT!

Intuitively think about two identical springs. One is at zero compression and the other is at 20cm compression.
Which do you *intuitively think* would be harder to compress further? The spring at rest (zero compression) or the spring that is already slightly compressed (20cm compression)?

If you think about this, I think that you'll realize that it's gets harder to compress spring further you push it. Thus, the equations for Energy and Force, have to represent the instantaneous values for any given compression length, and do not represent general values over the entire length of the spring.
Knowing that, you should realize that the equation 1/2kx^2 gives you the energy of the spring for only one specific compression length. So, to find the difference in energy between a 10cm compression and the 50cm compression you have to find the difference between the energies at those points. Or, as BRT already showed, you subtract 1/2kx1^2 - 1/2kx2^2.

 

[AZFutureDoc]

I remember making this VERY mistake earlier on in my studying. I will never make it again and always remember that you are not allowed to just add and subtract the displacements like the poster did. One of the reasons practice is so much better than looking at formulas.