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Work: an object falls onto a spring. Find max. dist. the spring is compressed.

  1. Oct 19, 2011 #1
    Until I learn to walk my name might pop up a lot in this thread...

    A block of mass m = 2.4 kg is dropped from height h = 59 cm onto a spring of spring constant k = 1240 N/m. Find the maximum distance the spring is compressed.

    Answer: 0.170m


    The first thing I did was convert the height from centimeters to meters so that all the units could work nicely together giving me 0.59 m.

    Then I found Work using mgrCos(theta).
    W = mgrCos(theta)
    W = 2.4 kg(9.8 m/(s^2))(0.59 m)Cos(0)
    W = 13.8768 J

    After I played around with a more general equation of Work.
    W = ΔK + ΔU
    W = Kf - Ki + Uf - Ui

    However, since the spring is in equilibrium before the block falls on it, conceptually, I saw that the Ui should equal to zero so my equation is,
    W = Kf - Ki +Uf

    And we know that Ki = 1240 N/m so,
    W = Kf - 1240 N/m + Uf
    W = 1/2m(v^2) - 1240 N/m + mgy

    13.8768 J = 1/2(2.4 kg)(Vf^2) - 1240 N/m + 2.4 kg(9.8 m/(s^2))(y)
    Now, I wrote Vf because it's an unknown and though I think there is a way to solve it using Work I went on to use a kinematic equation because I'm still very new to Work and have not yet learned how to do that.
    Vf^2 = Vi^2 +2aΔy
    Vf^2 = 9.8 m/(s^2)(0.59 m)
    Vf^2 = 11.564 m/s

    Back to my equation,
    13.8768 J = 1/2(2.4 kg)(11.564 m/2) - 1240 N/m + 23.52 N(y)
    13.8768 J = 13.8768 J - 1240 N/m + 23.52 N(y)
    0 = -1240 N/m + 23.52 N(y)
    1240 N/m = 23.52 N(y)
    52.72 /m = y
    _______________________________________________
    As you can see something went horribly wrong! I'm new to Work but I felt pretty confident approaching this problem until I got to the end. Not only is my answer numerically incorrect but the units seem impossible; nothing over meters?

    Can somebody help show me what I did wrong with this question? I'm sorry if it was difficult to follow, mathematics on text always seems like a puzzle to me.

    Thank you for taking the time to review my question, I greatly appreciate it.
     
  2. jcsd
  3. Oct 19, 2011 #2
    i think u r wrong at taking the height that the object falls as 0.59m. actually the height should be 0.59 +x, where x is the compression of the spring. at that point there is no KE. so, the relevant eqn. should be (from conservation of energy),
    mg(h+x) = 1/2 kx2.
    this is a quadratic eqn. in x. now solve for x, if u know how to solve a quadratic eqn.
     
  4. Oct 19, 2011 #3
    Thank you, you are absolutely right! The math is so much shorter this way too!
     
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