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Spring truck problem

  1. Apr 10, 2005 #1
    2. A spring from the truck is found not to conform to Hooke’s law. The force law for this spring is
    found to be:
    F = – k1x – k2 x^2
    where k1 = 52.8 N.m-1; k2 = 38.4 N.m-2.
    a) Compute the work required to stretch the spring from x = 0.500 m to x = 1.00 m.
    b) With one end of the spring fixed, a particle of mass 2.17 kg is attached to the other end of the
    spring when it is extended by an amount x = 1.00 m. If the particle is then released from rest,
    compute its speed at the instant the spring has returned to the configuration in which the extension
    is x = 0.500 m.
    c) Is the force exerted by the spring conservative or nonconservative? Explain your answer.

    Ok, for a)

    sub in the numbers for x = 1m and you get -91.2 N
    sub in the numbers for x= .5m and you get - 36 N

    So force is 127.2 N when the spring is stretched from .5 to 1m?

    As W = F*x, do i just plug in x=.5m and F = 127.2 N?

  2. jcsd
  3. Apr 10, 2005 #2
    In a and b. Calculate the energy in x = 0,5 and x = 1,00. You can't do a with the simple W = Fs because F is changeable.
  4. Apr 11, 2005 #3
    So you have to integrate?
  5. Apr 11, 2005 #4
    Yeah, I think so,

    Work done = Integral F.dx : Limits of Integraton are from x = 0.5 to x = 1.00
  6. Apr 11, 2005 #5
    Cheers :smile:
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