Springs and Hooke's law

In summary, the conversation discussed the use of Hooke's law to calculate the force and work required to stretch a spring. However, since the force is not constant as the spring is stretched, the method of multiplying the final force by displacement is incorrect. The concept of elastic potential energy was also introduced and used to calculate the work required in part (a) of the problem. In part (b), it was important to take into account the additional work needed to stretch the spring from 1 to 2 meters, rather than from 0 to 2 meters.
  • #1
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[SOLVED] Springs and Hooke's law

Homework Statement


An unstretched spring has a force constant of 1200 N/m. How large a force and how much work are required to stretch the spring by 1.00 m from its unstretched length?


Homework Equations



F= -k*s

W= F * s

The Attempt at a Solution



I used Hooke's law and obtained a force of 1200 N (which was correct). But the displacement is only 1 meter, so work should have also been 1200 (J). But it's wrong anyway.

The second part of the problem asks "How large a force and how much work are required to stretch the spring by 1.00 m beyond the length reached in part (a)?" so I multiplied the force constant by 2 to get 2400 N, and it was right. However, multiplying 2400 by 2 meters to give W of 4800 J was also incorrect. I don't know what I'm misunderstanding.
 
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  • #2
As you do work on the spring to stretch it, the force is not constant. So you can't just multiply the final force times the displacement. (You can integrate, if you know how.)

Hint: Have you studied elastic potential energy? How much energy is stored in a stretched spring?
 
  • #3
The course I'm in is *supposed* to be algebra and trig-based, but I am quickly learning that using calculus would make my life a lot simpler, if I only remembered how to do that stuff. We also haven't studied elastic potential energy yet... I looked on Wikipedia and it said this was found by integrating Hooke's law (which I would love to do if I knew how) and gave a formula. U = 1/2 (kx[tex]^{2}[/tex]) So I plugged in my values for the force constant and displacement, and got elastic potential energy of 600. I used this as my value for the work done in part a), and 600 J was correct.

However... I tried the same strategy on part b), by using the same force constant and a displacement of 2 meters, but the resulting 2400 J was also wrong. What did you mean by not being able to multiply final force by displacement?
 
  • #4
rolodexx said:
However... I tried the same strategy on part b), by using the same force constant and a displacement of 2 meters, but the resulting 2400 J was also wrong.
That's because the question asked for the additional work needed to stretch the spring from 1 to 2 meters, not from 0 to 2. (Subtract.)
What did you mean by not being able to multiply final force by displacement?
I was referring to your earlier idea for calculating work. For example, you multiplied 1200 N x 1 m, which is incorrect because the force actually varies from 0 to 1200 N as you stretch the spring.
 
  • #5
oh! Thank you so much... it definitely helps to "talk" to someone about this instead of just giving up. :smile:
 

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