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What should be the force constant of the spring?

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data
    You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1100 kg car moving at 0.66 m/s is to compress the spring no more than 9.0×10−2m before stopping.

    What should be the force constant of the spring? Assume that the spring has negligible mass.
    Express your answer using two significant figures.

    2. Relevant equations
    F=ma
    w=F*d
    w=ΛKE

    3. The attempt at a solution

    I Know that the KE o the car is transferred into the PE of the spring
    .5mv2=.5kx2
    Using this method I found the right answer = 59155.5

    But this question is in a chapter before PE is mentioned so Id like to see how else it can be solved. I initially tried this method, and would like to know why it is wrong:

    The car must decelerate during this displacement so using kinematics:
    02=662 + 2a(.09)
    -.4356=2a(.09)
    a= -2.42

    F=ma
    F=1100*-2.42
    F=-2695
    so the spring must apply the same force in the opposite direction to stop the cart
    29577=k(.09)
    k= 29577.8 (wrong answer)
    So why is this method not applicable to this problem?

    Also should integration be involved here somewhere? Where?
     
  2. jcsd
  3. Mar 24, 2016 #2

    mfb

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    Staff: Mentor

    The acceleration of the car and the force of the spring won't be constant, you cannot apply formulas that assume this.
     
  4. Mar 24, 2016 #3

    gneill

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    The acceleration is not constant for this problem since the force increases as the spring compresses. The formula that you've chosen is one of the SUVAT equations which only apply when the acceleration is constant.

    Two of your relevant equations involve work. Are those equations fair game for solving the problem?
     
  5. Mar 24, 2016 #4
    Well, the formula for work, KE_f - KE_i was derived by plugging one of the SUVAT equations into W=m*a*d
    But, that equation can also be found by integration, so does Change in KE hold even when acceleration is not constant?

    So how would I solve this problem not using spring PE and using integration?
     
  6. Mar 24, 2016 #5
    you have taken average deceleration 'a' throughout the motion but the the spring force builds up during its compression -its proportional to change in length- so 'a' will vary.
     
  7. Mar 24, 2016 #6

    gneill

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    Yes, but you need to use a method that takes the changing acceleration into account, or bypasses it altogether (such as using conservation of energy and the change in PE).
    You know the expression for the force for a given distance of compression of the spring (Hooke's Law). Look at integrating F*d to find the work done; it's one of your relevant equations.
     
  8. Mar 24, 2016 #7
    integrating hooks law i get :
    ∫kx
    .5kx2

    what can I do with this?
     
  9. Mar 24, 2016 #8

    gneill

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    Does it remind you of an energy formula that you used for your first attempt? :wink:
     
  10. Mar 24, 2016 #9
    Hm, the potential energy of the spring. This represents the work since the integration is of Force over dl, so I set that to equal the negative work the car does on the spring and I end up with .5mv2=.5kx2 again. So it seems this is the only way to solve it?
     
  11. Mar 24, 2016 #10

    gneill

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    You may be able to write a differential equation for the motion and solve it. But that would probably be more effort than is warranted.
     
  12. Mar 24, 2016 #11
    I prefer work energy theorem over differential equations. Can I ask for assistance in one more question very closely related to this in this thread?
     
  13. Mar 24, 2016 #12

    gneill

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    Staff: Mentor

    You can ask. Once I see your question I'll tell you if it needs a separate thread.

    Edit: After seeing the new material I decided that a new thread was in order.
     
    Last edited: Mar 24, 2016
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