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amistre64 Group Title

Object sliding down a ramp .... PE and KE

  • 2 years ago
  • 2 years ago

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  1. amistre64 Group Title
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    |dw:1348751150306:dw|

    • 2 years ago
  2. amistre64 Group Title
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    oh, and time it takes to slide to the bottom of the ramp is 1.638 seconds Potential energy = .0025*9.8*.46 Kinetic energy at the bottom of the ramp is?

    • 2 years ago
  3. amistre64 Group Title
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    this problem involves friction

    • 2 years ago
  4. imron07 Group Title
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    I think we can use work-kinetic energy theorem.

    • 2 years ago
  5. amistre64 Group Title
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    hopefully :) I seem to come up with 2 different answers; and i know one of them is wrong.

    • 2 years ago
  6. imron07 Group Title
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    The aceleration must be constant right?

    • 2 years ago
  7. amistre64 Group Title
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    \[a=\frac{2d}{t^2}\] \[v=at=\frac{2d}{t}\] \[K_e=\frac12mv^2\] energy lost to friction \[P_e-K_e\]

    • 2 years ago
  8. amistre64 Group Title
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    assume a constant acceleration yes

    • 2 years ago
  9. imron07 Group Title
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    You're did it correct i think

    • 2 years ago
  10. rajathsbhat Group Title
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    do you know the co-efficient of sliding friction?

    • 2 years ago
  11. amistre64 Group Title
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    not outright; but that can be calculated with given information i believe

    • 2 years ago
  12. imron07 Group Title
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    Hmm, actually we don't need \(\mu\) since time is known.

    • 2 years ago
  13. rajathsbhat Group Title
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    mmm true.

    • 2 years ago
  14. amistre64 Group Title
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    \[\mu_k=\frac{g~sin\theta-a}{g~cos\theta}\] \[F_k=\mu_k*\vec n=\frac{g~sin\theta-a}{g~cos\theta}*mg~cos\theta\] \[F_k=m(g~sin\theta-a)\] \[W_f=F_k*d=md~(g~sin\theta-a)\]

    • 2 years ago
  15. rajathsbhat Group Title
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    hey you know that v=2d/t^2. So KE is just 1/2 mv^2.

    • 2 years ago
  16. rajathsbhat Group Title
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    i mean, what else do you want?

    • 2 years ago
  17. amistre64 Group Title
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    just working it thru a couple of ways. When I first tried to work this out; I used: Pe = mgh = 0.01127 J in a frictionless system; Ke = Pe at the bottom of the ramp mgh = mv^2/2 gh = v^2/2 sqrt(2gh) = v v = 3.0027 m/s right?

    • 2 years ago
  18. imron07 Group Title
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    The problem maybe 1.638 sec isn't the correct time :)

    • 2 years ago
  19. rajathsbhat Group Title
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    yes, v=3m/s in that case.

    • 2 years ago
  20. amistre64 Group Title
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    its pretty good, we slid the penny 5 times and averaged it out

    • 2 years ago
  21. amistre64 Group Title
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    at first the velocity of the Ke i just worked out threw me for a loop

    • 2 years ago
  22. rajathsbhat Group Title
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    you actually measured 3m/s!?

    • 2 years ago
  23. amistre64 Group Title
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    i couldnt see why the velocity from v=2d/t wasnt matching up :)

    • 2 years ago
  24. amistre64 Group Title
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    the 3 m/s was not measured, it was calculated

    • 2 years ago
  25. rajathsbhat Group Title
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    oh i thought you performed an experiment with a penny..

    • 2 years ago
  26. amistre64 Group Title
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    i spose my question is; is using the velocity of 2d/t a proper way to determine the kinetic energy in the system at the bottom of the ramp?

    • 2 years ago
  27. rajathsbhat Group Title
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    yes. This works no matter what the friction is (even 0).

    • 2 years ago
  28. amistre64 Group Title
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    thanks :) so if it was a frictionless system, then the kinetic energy at the bottom of the ramp would still equal the potential energy from the start of it; and all that would change is the final velocity of the object then.

    • 2 years ago
  29. amistre64 Group Title
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    ..and a change in time, but thats a given :)

    • 2 years ago
  30. rajathsbhat Group Title
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    yes.

    • 2 years ago
  31. rajathsbhat Group Title
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    Here's a nice video of an extreme case where the object slides with uniform velocity: http://www.youtube.com/watch?v=84DXjumAxNQ&feature=plcp

    • 2 years ago
  32. amistre64 Group Title
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    lol, great; i minute ten watching a sock slide down a banister ....

    • 2 years ago
  33. rajathsbhat Group Title
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    x)

    • 2 years ago
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