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Energy Problems (Help Greatly Needed!)

  1. Oct 24, 2004 #1
    I am having some difficulty answering a few problems on my physics homework.
    I don't think that I am coming up with the right equations for given situations in the problems. The questions are as follows:

    1. Question 31

    For this question I am getting [tex]- \frac{1}{2} m_1v^2 - \frac{1}{2} m_1v^2+m_2g\Delta h=-m_1\mu_kd[/tex] where m1 is the block in contact with the surface and m2 is the hanging sphere. When I solve the equation for v (velocity), I get v=4.77 m/s. However, I should be getting a result of 3.74 m/s.

    2. Question 48

    For this equation I get [tex]mgy_{max}-mgh=-mg \mu_k d[/tex] but there needs to be something with the angle to derive that equation given in the problem.

    3. Question 60
    Accompanying Figure

    When doing part (a) I was using the equation [tex]\frac{1}{2} m v^2+\frac{1}{2} k d^2=0[/tex]to find d. I got d=0.424 m but for part (b) I would just get v=3 m/s which is given in the question when the object is moving to the right. I'm not sure if this is right or if I'm approaching the question correctly.

    4. Question 72

    For this question I was able to find part (b) but I'm not sure how to start in finding part (a).

    Any Help is greatly appreciated.
    Last edited: Oct 24, 2004
  2. jcsd
  3. Oct 24, 2004 #2


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    Homework Helper


    You should have

    [tex] \frac{1}{2}m_{1}v^2 + \frac{1}{2}m_{2}v^2 - m_{2}gh = -\mu m_{1}gh [/tex]

    You should have

    [tex] mgy_{max} - mgh = -\mu mgcos\theta \frac{y_{max}}{sin\theta} [/tex]

    3) Read the problem again you seem to forgot friction on this problem
    Last edited: Oct 24, 2004
  4. Oct 24, 2004 #3
    I know that there is friction but when does it need to be accounted for?

    If I were to use [tex]\frac{1}{2} mv^2+\frac{1}{2} kd^2=-\mu_kmgd[/tex] I would have two variables to solve for because I do not know what the distance the block is sliding before it reaches the spring. Or are you saying that I need to use this equation for the other parts of the question if that is the right equation.
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