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Last 2 forces questions

  1. Jul 1, 2003 #1
    http://cummingsiam.steven-sst.com/argh.jpg [Broken]

    question 8 is similar to my last forces question. I got 3.2mm displacement for the raquet - yet the book says 2.2
    The only difference i see is that the ball rebounds at 9.5ms-1 and instead of compression - we have displacement. Its the same thing...

    question 9 is damnright confusing..
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Jul 1, 2003 #2


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    How did you go about getting 3.2mm? Please post your work so we can check where your (or the book's) mistake is.

    For the second one, remember Work = Force * distance. You'll need to do a few simple integrations.
  4. Jul 2, 2003 #3

    the speed the ball is traveling is 10ms-1
    the rule for kinetic energy is .5 * mass * velocity squared.
    so .5 * .1 * 100
    = 5j of energy

    now, work done is measured in jouls.. and can be measured as the area under the force displacement graph.
    so 5j = .5 * force * displacement.
    if we let X = force and displacement
    5j = .5Xsquared
    10 = X squared
    X = 3.2 mm displacement.

    i let X = both force and displacement as if i change the scale of the force..X can equal both sides.

    i get 3.2mm..the book says 2.2
  5. Jul 3, 2003 #4


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    "if we let X = force and displacement"????

    I don't know what you mean by "scale the force". What scale are you using? That should depend on the "spring constant"- the slope of the Force versus displacement graph. Unfortunately, I was unable to open your link to find that. What is the slope of the graph?
    You should be able to write the force as k*X, reading k from the graph.
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