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W=(1/2)KA^2 Can someone explain this equation in detail & Q

  1. May 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass resting on a horizontal frictionless surface is attached to one end of a spring. 3.6 J to compress spring 0.13 m. released at max acceleration 15 m/s. What is the spring constant and mass

    I would like to dicuss this question how the equations are derived. can someone give me there explainantion of the problem attached?
    2. Relevant equations
    W=(1/2)KA^2

    3. The attempt at a solution
    426 n/m and 3.7 kg
     

    Attached Files:

  2. jcsd
  3. May 3, 2016 #2
    Try to derive it yourself. Ill give you a hint. Restoring force applied by a spring is directly proportional to its contraction/expansion in the direction to attain natural length. F=-Kx, also work done is F. dx. Try getting an answer
     
  4. May 3, 2016 #3
    I know using Hookes law can be used after finding the mass. I want to know more how W=1/2KA^2 is derived in this solution. My understanding is W is the potential energy the and A^2 is the amplitude and the reason it is square because ????
     
  5. May 3, 2016 #4

    Ray Vickson

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    You have an expression for the force ##F## exerted by the spring when it is compressed to length ##x## (Hooke's Law). How would you compute the work needed to compress the spring from its natural length to a smaller length?

    Note: you should make a serious attempt to understand what is being asked here, and how to do it.
     
  6. May 3, 2016 #5
    Well, whatever force we apply on the spring to contract it is stored in the spring as its potential energy. So the work done in compressing the spring is F.x. Where x is the compression in the spring. As the force varies with the compression(kx) here we calculate the work done for elemenal compression dx. Work done dw=potential of spring =F. dx=-Kx.dx.

    As the force is opposite to the elemental compression dW=Kxdx.
    Integrating with limits from 0 to x. We get W=integration(kxdx) =1/2Kx^2.
     
  7. May 3, 2016 #6

    Ray Vickson

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    Good.
     
  8. May 3, 2016 #7

    BvU

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    Ray,

    I have the impresssion Abhi made a good start inviting OP to find out for himself and then spoilt the fun by providing the answer !

    PB: is it clear to you now ?
     
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