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Homework Help: Hooke's Law and a spring

  1. Jul 26, 2006 #1
    Okay, I'm having a debate with my teacher. He's saying I'm wrong but I still think I'm right. The question is:

    What should be the spring constant, k, of a spring designed to bring a 1200kg car to rest from a speed of 100km/h so that the occupants undergo a maximum acceleration of 5.0g?

    He said to use the equations v2^2 = v1^2 + 2ad and 1/2mv2^2=1/2mv1^2+mad

    I said ma=kx, ma/k=x

    then 1/2mv^2=1/2k(ma/k)^2, solve for k and I get 3700, whereas he got 14,000. Could someone please help me out with this?
     
  2. jcsd
  3. Jul 26, 2006 #2
    Infact, I get a value of just over 14,000 using your method. Did you convert the speed to m/s? Also use g = 9.8 ms-2.
     
  4. Jul 26, 2006 #3
    I used g, yes. and I did convert.
     
  5. Jul 26, 2006 #4
    I'm really sorry, I actually get a value of just over 3700.
     
  6. Jul 26, 2006 #5
    What formula is that?

    I would do it differently. :tongue2:

    First, I used this formula do find the displacement of the spring:

    [tex]v^2 = {v_0}^2 + 2ax[/tex]

    It gives [tex]x = 8.0[/tex] m.

    Then, I used Hooke's Law to find the spring constant.

    [tex]F = -k.x[/tex]

    [tex]m.a = -k.x[/tex]

    [tex]k = \frac{-m.a}{x}[/tex]

    [tex]k = \frac{-1200 \times -5 \times 9.80}{8}[/tex]

    [tex]k = 7350[/tex] N/m
     
  7. Jul 26, 2006 #6
    Yes, I did too!

    See, my reasoning is, F=ma=kx, but k and m are constant, therefore when a greater force is applied to scretch the spring (or in this case, compress), acceleration changes (as well as x). However, the other two equations involves a uniform acceleration, which you can't have in this case.
     
  8. Jul 26, 2006 #7
    But in this case, isn't the car undergoing a constant acceleration of -5g? :grumpy:
     
    Last edited by a moderator: Jul 26, 2006
  9. Jul 26, 2006 #8

    Andrew Mason

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    You cannot use [itex]1/2mv2^2=1/2mv1^2+mad[/itex] because the acceleration is not constant.

    If the maximum acceleration is 5 g = -49 m/sec^2 then kd = 5mg where d is the stopping distance. So k = 5mg/d

    The work done in stopping the car is stored in spring: [itex]\frac{1}{2}mv^2 = \frac{1}{2}kd^2[/itex], so:

    [tex]kd^2 = 5mgd = mv^2[/tex]

    [tex]5gd = v^2[/tex]

    [tex]d = v^2/5g = 15.75 m[/tex]

    So k = 5mg/d = 5*1200*9.8/15.75 = 3700 N/m to two significant figures.

    AM
     
    Last edited: Jul 26, 2006
  10. Jul 26, 2006 #9
    Ah... of course it is not constant. Sorry, my mistake.
     
  11. Jul 26, 2006 #10
    It's a maximum acceleration
     
  12. Jul 26, 2006 #11

    nrqed

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    I can't check the numbers right now but your method seems correct. And of course, as you pointed out, the equation he suggested is wrong because the acceleration is not constant. I am sure that when you tell him/her that, he will slap himself and agree with you!
     
  13. Jul 26, 2006 #12
    Yes, know I understand. I was naive. :grumpy:

    Thank Hollysmoke and Andrew Mason! :approve:
     
  14. Jul 26, 2006 #13
    Actually, that's why I'm here. I told him and he brushed me off, saying F=ma=kx <- Acceleration is constant. I actually just stormed out of the class and had a discussion with the VP about it. I just wanted to make sure I wasn't making a total ass of myself though. Thanks for your help :3
     
  15. Jul 26, 2006 #14

    Andrew Mason

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    Ask him how the force can be constant if x changes. If he says it doesn't change, ask him to explain physically how the spring slows the car down. Perhaps there we have a different understanding of how the spring is configured.

    AM
     
  16. Jul 26, 2006 #15
    Okay, I'll do that. I'll tell you how it goes tomorrow ^^
     
  17. Jul 26, 2006 #16

    nrqed

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    :surprised :surprised :surprised :surprised
    I am flabbergasted!!
    I mean, I can imagine a prof being distracted and for a second using an invalid equation, but to *defend* that the acceleration is constant for mass connected to a spring is amazing. Wow...

    Good luck!
     
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