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Conservation of energy

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    equation in the problem

    k = 898 N/m
    m = 90.7g = .0907kg
    d = 27 inch = .6558m
    g = 9.8 m/s

    3. The attempt at a solution

    h = [(898n/m)/(2)(.0907kg)(9.8m/s)](.6558m)^2 - (.6558)

    that is for the H. when computing this is got 217.55m. I am assuming the way the equation is written out for h in the problem that is how to set it up.

    now it asks for velocity.

    when it gets back to zero. We will say E1 = E2....

    at E1 it is all potential energy and no kinetic. and at E2 where we find final velocity it is all kenetic and no potential?

    so then we have mgh = 1/2 mv^2......

    plus the top equation in for h

    then isolate V and solve.

    thanks for any help.

    Attached Files:

  2. jcsd
  3. Nov 12, 2008 #2
    just curious do most people not like attachments....

    Its the best way for me to post the problems because a lot of equations have trouble making readable just typing in.
  4. Nov 12, 2008 #3
    Here's his problem:

    6.) A sphere of mass [tex]m[/tex] hangs from a string of length [tex]l[/tex] and is drawn back so that the string makes an angle of [tex]\theta[/tex] with the vertical axis. It swings down until the strings hits a peg which is [tex]l/2[/tex] directly below the point where the string is attached to the ceiling. When the string hits the peg, the sphere keeps moving and swings up through an angle [tex]\alpha[/tex]. Use conservation of energy to find an equation for the angle [tex]\alpha[/tex]. Show that this equation makes sense with a few, well-chosen examples for [tex]\theta[/tex].
    Last edited: Nov 12, 2008
  5. Nov 12, 2008 #4
    sent. ty.
  6. Nov 12, 2008 #5
    The problem's posted in the above post; if you want, you can "quote" me and select what I typed and put it in the original post. Otherwise, here's my suggestion:

    You know the conservation of energy equation:

    [tex]U_0+K_0=U+K[/tex], where [tex]U[/tex] indicates potential energy and [tex]K[/tex] represents kinetic energy. What do the two critical moments have for their energy?
  7. Nov 12, 2008 #6
    yes i am sorry that one is the wrong question for the work i included...resent you the right one. I will most likely be able to use that one too though so thank you.
  8. Nov 12, 2008 #7
    Neglect air resistance.

    The height of an arrow shot directly upwards from an inital height [tex]0m[/tex] is given by: [tex]h=\frac{k}{2mg}d^2-d[/tex]. [tex]k=898N\cdot m[/tex], mass is [tex]0.0907kg[/tex], and [tex]d=27in[/tex]; find [tex]h[/tex]. And, find [tex]v[/tex] at [tex]h=0[/tex].
  9. Nov 12, 2008 #8
    ok put the two together
  10. Nov 12, 2008 #9
    That sounds perfect.
  11. Nov 12, 2008 #10
  12. Nov 12, 2008 #11

    original problem
  13. Nov 12, 2008 #12
    my solution for the first part.

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