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Homework Help: Potential and Kinetic Energy: Down an incline with spring

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data

    A 4.1 kg block starts at rest and slides a distance d down a frictionless 26.0 deg incline, where it runs into a spring. The block slides an additional 18.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 428 N/m .

    [URL]http://loncapa4.fsu.edu/enc/64/444d8fc39f68b5d50a0ae69506afa18b07b17f4630544f446b909dc06ad62e4ccad64d37e395dbf1b8bce7f718c3ca893e7291ecae8bce02527ece77d5a672e050dafc3a7d01eb0880c83b19b1ecf1e46bc159a279e62c69.gif[/URL]

    What is the value of d?..(answered)

    What is the distance between the point of first contact and the point where the block's speed is greatest?.....(need help)



    2. Relevant equations

    U = mgh

    U = .5k(x^2)

    3. The attempt at a solution

    I found the answer to the first question by using the above relevant equations:

    U = mgh = .5k(x^2)

    4.1 * 9.81 * h = .5 * 428 * (.18^2)

    h = 0.172 m

    then some trig..

    sin(26)=.172/(.18 + d)

    d = .213 m

    As far as how to solve the second question I am completely lost. At the point of contact with the spring the acceleration should be zero correct. From there would I need to find the intial velocity? We know that V_f = 0, but how to use these "known" quantities in relationship to the block being slowed by the spring is confusing me. I think I'm making this problem more complicated than it realy is. TIA for your help.
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Sep 27, 2010 #2
    I tried using:

    mgh = F_net * delta(x)

    0.172 = (mgsin() - .5k(x^2))(.18 - x)

    x = .162 m

    This approach did not work...
     
  4. Sep 28, 2010 #3
    Ok, I found with some help that mgsin()=kx' will get me the correct answer....however I am still confused where this came from....
     
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