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Mechanical conservation of energy

  1. Nov 9, 2007 #1
    A 2.5-kg object suspended from the ceiling by a string that has a length of 2.5 m is released from rest with the string 40 below the horizontal position. What is the tension in the string at the instant the object passes through its lowest position?

    how do I find the velocity using mechanical conservation of energy the angle is making me confuse?
     
  2. jcsd
  3. Nov 9, 2007 #2
    Can you begin by writing the equation for the conservation of mechanical energy?
     
  4. Nov 9, 2007 #3
    the equation will be mgh = 1/2mv^2

    the masses wil cancel out but I am not sure how to incorporate the angle

    gh = 1/2 * sin(40)^2 ??
    I think is wrong
     
  5. Nov 9, 2007 #4
    Just so we're on the same page, you've decided that the ball at the bottom of the swing will have zero potential energy, and its initial kinetic energy is also zero. You've found part of the height, in other words, we're not going to use 2.5sin(40) for the height, we're interested in the balls height from the point of zero potential energy. I'm not seeing exactly how you're plugging in the numbers, it looks like you placed the value of the initial height into the speed variable.

    From what you wrote, you should have the equation: [tex] v_f = \sqrt{2gh_0} [/tex]

    Let's work on finding h initial, whats the relation of 2.5sin(40) to the final height? Try drawing this picture to see what I mean.
     
  6. Nov 9, 2007 #5
    the final height will be 2.5 so 2.5sin40 is part of the height therefore to find the height we go down I need to substract 2.5 - 2.5sin40 ????
     
  7. Nov 9, 2007 #6
    Well, 2.5 isn't the final height, zero would be the final height. This is what allowed you to make the final potential energy of the ball zero. But yes, the initial height is 2.5 - 2.5sin40. Let me know if you get hung up.
     
  8. Nov 9, 2007 #7
    i got it thank you very much V=4.17 and T=41.8
     
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