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Potential energy questions

  1. May 28, 2006 #1
    Hello,
    This is my first post, so please forgive any mistakes I might have made. I am having a little difficulty with these 3 problems. Any push would be much apperciated. Thanks!

    1. A 2.0 kg block is dropped from a height of 80 cm onto a spring of force constant k = 1960 N/m, as shown in Fig. 12-19. Find the maximum distance the spring will be compressed.
    I figured that energy due to gravity must equal the energy of the spring - in other words, ky^2/2 = mgy.

    2.Tarzan, who weighs 655 N, swings from a cliff at the end of a convenient vine that is 21 m long (Fig. 12-21). From the top of the cliff to the bottom of the swing, he descends by 3.2 m. The vine will break if the force on it exceeds 950 N. What is the greatest force on the vine during the swing?
    I figure that greatest force will be when the force of gravity adds up with the force of the circuluar motion. However, I do not believe there is enough information to compute the velocity of the swing.
     
  2. jcsd
  3. May 28, 2006 #2

    Hootenanny

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    Hi there James and welcome to PF,

    You're answer to question one is spot on, just a minor point however you stated the equation as 1/2ky2 = mgy; this is not correct because the compression of the spring will not equal the height from which the block was dropped, you're equation should be 1/2kx2 = mgy where x is the compression. As for question two, there most certainly is enough information to calculate the velocity at the lowest point, which is where the tension will be greatest. Think about potential energy.

    ~H
     
    Last edited: May 28, 2006
  4. May 28, 2006 #3
    Just wondering, why would it be kx^2 since the work done if the spring is stretch or compressed is = 1/2(k)(x)^2 which is the gain in kinetic energy of the block from the drop?
     
  5. May 28, 2006 #4

    Hootenanny

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    Ahh, sorry thank you for the correction al. I just missed the half off by accident :blushing: . I have duely recitified my error.

    ~H
     
    Last edited: May 28, 2006
  6. May 30, 2006 #5
    On the first one, I've tried solving for the equation 1/2ky2 = mgy, which worked out to be x = sqrt(2mgy/k). Plugged in all the numbers sqrt((2*2*9.8*.8)/1960), and I got .12649, which did not work.
     
  7. May 30, 2006 #6

    Hootenanny

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    I apologise, I forgot to add on the additional potential energy loss due to the spring compressing. The formula should be;

    [tex]\frac{1}{2}kx^2 = mg(h+x)[/tex]

    The block will lose additional potential energy when the spring is compressed. If you expand the above expression out, you should obtain a quadratic equation in terms of x.

    Apologies for the mix up. Do you undersand where the formula come from? The (h+x) is the height from which the ball was dropped plus compression of the string, which represents the total loss of potential energy.

    ~H
     
    Last edited: May 30, 2006
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