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I Need Physics Help Bad

  1. Nov 21, 2005 #1
    Can someone please tell me how to do these problems? Show work if possible please.
    I'm not sure at all how force is tied in to Potential Energy so I'm not getting anywhere on the first problem...
    Last edited: Nov 21, 2005
  2. jcsd
  3. Nov 21, 2005 #2


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    Shouldn't that be what we here would say to you? Psst, forum rules.
  4. Nov 21, 2005 #3
    Sorry, I just need at least help starting. Physics is blowing my mind.
  5. Nov 21, 2005 #4
    Ok, I'm still totally stuck but I think F = mv^2 / r. I'm not sure how to incorporate the other variables... I'm not exactly sure how Potential Energy can be replaced with force... so I'm stuck

    I think I need to mix in Kinetic Energy somewhere... but I'm not positive at all where. That's pretty much as far as I can get...
  6. Nov 21, 2005 #5


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    For the first question, this is a conservation of energy problem. The total energy (potential mgh + kinetic 1/2mv^2) remains constant. Depending on how much potential you have at each point, you can therefore calculate the corresponding kinetic energy.

    Knowing kinetic energy means knowing velocity if mass is given. There is a direct relationship between centripetal force and velocity.
  7. Nov 21, 2005 #6
    Omg thanks so much. So mgh +1/2mv^2 = E net?
    Or is it mgh + 1/2mv^2 = mg(2r) + 1/2mv^2?
  8. Nov 21, 2005 #7


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    What I said about the total energy being constant is that you first find a point where BOTH the potential and kinetic energies are known so you may evaluate this constant. We should choose the point where it is at the top of the slope, then KE = 0, and PE = mgh. So we now know the total energy of the system. The total energy anywhere else in the system must be equal to this.

    At the point at the top of the circle, you have mg(2r) + 1/2mv^2 = mgh as you stated ^^
  9. Nov 21, 2005 #8
    That makes sense. If mg(2r) + 1/2mv^2 = mgh represents the total energy in terms of the top of the circle, would mg(2r) = mgh - 1/2mv^2 represent the energy at the bottom?

    Actually I dunno about that previous statement...could I solve for v = (2gh - 2g(2r))^(1/2) and then plug that into F = mv^2/r?
  10. Nov 21, 2005 #9


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    Apparently, you are still jumbling with the equations without much understanding. We should always discuss physics using concepts, then use equations to quantify them.

    You know the total energy of the system (PE + KE) at any point is always mgh. Because of the given information, you always know the potential energy, thus kinetic energy can always be found. Kinetic energy is described by mass and velocity. If you know one, you can always find the other right?

    If you want to find the kinetic energy at the bottom of the ramp, what are the conditions for the PE and KE?

    If you have solved legitimately for v using the conservation of energy concepts, there is no reason you cannot use it in other calculations.
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