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Change in Potential Energy

  1. Feb 27, 2014 #1
    1. The problem statement, all variables and given/known data
    A 20 kg child slides down a slide at a playground. The slide is 5 meters long, and is inclined at an angle of 37 degrees above the horizontal. What is the child's change in potential?

    2. Relevant equations
    PE=mgy

    With m=mass, g=gravity, and y=distance.


    3. The attempt at a solution

    I have solved potential energy problems before this,but I don't know how to account for the angle. If there was no angle, the formula would be PE=20*9.8*5, and I think that would be the change in energy since when the child reaches the bottom y=0, canceling out the entire equation. Could someone please tell me where the angle comes in? I don't want it done for me, I just need one post with a helpful explanation, and then I should be good. :)

    Thanks so much!
     
  2. jcsd
  3. Feb 27, 2014 #2
    There are two ways of doing this that give the same answer.

    Method 1: You use the equation you presented, but instead of the 5, you use the vertical distance that the child descended.

    Method 2. You use the equation you presented, but instead of the 9.8, you use the tangential component of g along the slide.

    Chet
     
  4. Feb 27, 2014 #3
    Thank you! I knew trig would come in somewhere, I just couldn't figure out how to bring it in. So using Method 1, would the vertical distance be 3.94 meters?
     
  5. Feb 27, 2014 #4
    No. Try again. Draw a diagram this time.

    Chet
     
  6. Feb 27, 2014 #5
    I drew one, but I did the trig wrong, I think. Would 5tan38 work?

    Thanks again!
     
  7. Feb 27, 2014 #6
    No. The hypotenuse is the slide. try again.
     
  8. Feb 28, 2014 #7

    adjacent

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    Gold Member

    attachment.php?attachmentid=67097&stc=1&d=1393579060.png

    Find h.
    You won't need me here(Adjacent side) :cry:
     

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  9. Feb 28, 2014 #8
    Hm, I'm not sure why I can't wrap my mind around this. It would have to be sine, but why? Wouldn't that be the two unknown sides?

    Thanks!
     
  10. Feb 28, 2014 #9

    Doc Al

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    Staff: Mentor

    What's the definition of sine for a right triangle?
     
  11. Feb 28, 2014 #10

    Doc Al

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    Staff: Mentor

  12. Feb 28, 2014 #11
    Sine=opposite/hypotenuse

    Oh! Obviously, 5 is the hypotenuse. I let adjacent's letter choice confuse me. So 5sin 37=3.0 m. Correct?
     
  13. Feb 28, 2014 #12
    Thanks, I know the functions fairly well, but I tend to overcomplicate them. Oops.
     
  14. Feb 28, 2014 #13

    Doc Al

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    Staff: Mentor

    Correct.
     
  15. Feb 28, 2014 #14
    PE=mgy
    PE=20*9.8*3.0=588 Joules

    Correct?

    Thanks again!
     
  16. Feb 28, 2014 #15

    Doc Al

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    Good. That's the magnitude of the change, but what's the sign of the change?
     
  17. Mar 28, 2014 #16
    I forgot about this thread. Oops. The sign, as in negative or positive? It depends on what direction we decided is the positive direction, right? So if we decided that down was the positive direction, then h would be negative, so the answer would be negative? But can you have negative energy? Would I just say 588 Joules down?

    Thanks!
     
  18. Mar 28, 2014 #17

    Doc Al

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    Right.

    It's got nothing to do with direction (energy is not a vector). It has to do with whether the potential energy increases (a positive change) or decreases (a negative change).

    Since the child slides down the incline, is her change in PE positive or negative?
     
  19. Apr 5, 2014 #18
    Her change in PE would be negative, so the KE would be positive.
     
  20. Apr 6, 2014 #19

    Doc Al

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    Right. Since the change in PE is negative (the PE decreases), the change in KE will be positive (the KE increases). (The KE itself is always positive--unless it's zero.)
     
  21. Apr 6, 2014 #20
    Thank you!
     
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