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Homework Help: First post: work and energy problem

  1. Dec 25, 2017 #1
    1. The problem statement, all variables and given/known data
    A 20.0 kg block on a horizontal surface is attached to a horizontal spring of k = 2.0 kN/m. The block is pulled to the right so that the spring is extended 10.0 cm beyond its unstretched length, and the block is then released from rest. The frictional force between the sliding block and the surface has magnitude 80.0 N.

    2. Relevant equations
    What is the maximum KE attained by the block as it slides from release to the point where the spring is unstretched?

    3. The attempt at a solution
    KE is maximized when velocity is maximized. Therefore, the point is when the two forces are at equilibrium.
    Set kx = Fd, d = (0.1-x)
    2000(x) = 80(0.1-x)
    x = 8/2080 (m)
    = 0.003846 (m)

    EPE (when fully stretched to 0.1 m) + Work done by friction = KE + EPE2
    (.5)(2000)(0.1^2) - 80(0.1-0.003846) = .5(2000)(0.003846^2) + KE
    Therefore, KE = 2.29 J, which makes sense.

    I also did it the calculus way
    W = the integral of force
    F = ma, m =20 kg
    a = [kx-f(0.1-x)]/20, integrating this yields velocity function, c =0
    derive that function and set it to zero, yields the same answer.

    But 2.29 J is not right. Why?
  2. jcsd
  3. Dec 25, 2017 #2


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    What are you doing here? Compare the units on either side of this equation!

    In general, always check that the units on each side of your equations match and never forget to write out the units.
  4. Dec 25, 2017 #3


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    Hi lanardo and welcome to PF.

    Your starting equations are incorrect.
    This is dimensionally incorrect because kx has dimensions of [force] and Fd has dimensions of [Force×distance].
    This is also incorrect for the same reason.
    Do the problem by applying the work-energy theorem when the spring is at some distance x from its unstretched position and the mass is moving with speed v. Use symbols and put in the numbers at the very end so you can see what's going on.
  5. Dec 25, 2017 #4
    I know what you mean. But how would you solve it if you were to not equate the two forces?
  6. Dec 25, 2017 #5
    Ok I see, thanks for the words!
  7. Dec 25, 2017 #6
    Got it! Thanks for both of your help!

    3.6 J is the final answer. I used the work-energy theorem and took the derivative AND set it to zero, find x. Then plug the x back in to solve for KE.

    Thank you for your advice. Merry Christmas!
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