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Physics speed and friction

  1. Jul 19, 2012 #1
    Question:

    557019_10151925919585526_976395695_n.jpg

    answers so far ( i dont know if its right thats why im posting on here)
    555686_10151925919215526_1057833896_n.jpg


    please let me know if im doing it correctly thanks !!!!! :smile:
     
  2. jcsd
  3. Jul 19, 2012 #2
    also i have no idea how to do c! at all !
     
  4. Jul 19, 2012 #3

    PhanthomJay

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    I don't know where your 'four meters' is coming from. [itex]KE_i + PE_i = KE_f + PE_f [/itex] , or [itex]\Delta KE + \Delta PE = 0 [/itex] if you like. Please explain.

    For part c, I don't see the rest of the pic after B. If it starts at 2.5 m, though, above the reference plane, where must it end up before it stops, without friction, using the same equation?
    .
     
  5. Jul 19, 2012 #4
    Don't you add 1.5 and 2.5 together ?
     
  6. Jul 19, 2012 #5
    I have not attempted c and need to redo the beginning part. So what do I out as my delta h value then? Is everything else in that equation correct?
     
  7. Jul 19, 2012 #6

    PhanthomJay

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    You are misreading the graph, and setting up your equation incorrectly as well.
    The skater starts off with an initial speed of 2.6 m/s^2 at a height 2 .5 m above a reference plane. For part a, at point A, the skater is 1.5 m above ground at some speed. The delta h is -1.0 m. But be consistent in setting up the energy equation. Either use initial KE plus initial PE = final KE plus final PE, where the initial PE is 2.5mg and the final PE is 1.5mg, or use the "delta" formula I referenced, where the change in PE is (1.5mg - 2.5 mg) = -1.0mg.
     
  8. Jul 19, 2012 #7
    The way i did it above is how my text book told me to do it. I can take a picture and post it of you would liek. Of an example. Because I am very lost now. So I can use hate equation I just did above ?
     
  9. Jul 19, 2012 #8

    PhanthomJay

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    Please use the first equation I gave you: 1/2m(2.6)^2 + mg(2.5) = 1/2mv^2 + mg(1.5). Solve for v at point A. Continue... Don't confuse h with delta h
     
  10. Jul 19, 2012 #9
    1/2(74.5)(2.6)^2 + (74.5)(9.8m\s)(2.5) = 1/2(74.5)v^2 + (74.5)(9.8m\s)(1.5)


    so i solve for v^2 like i normally would with the above equation ?
     
  11. Jul 19, 2012 #10

    PhanthomJay

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  12. Jul 19, 2012 #11
    and do the same for b... but replace with 1.5?
     
  13. Jul 19, 2012 #12

    PhanthomJay

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    No, start at the top again for initial condition (v = 2.6 and h_top = 2.5) and the final condition is at B (v = ? and h_B =0), which you sort of indicated the first time except you called it delta h instead of h_B, h_B is the height at B above the reference plane and is = to 0.
     
  14. Jul 19, 2012 #13
    So the value will be 0 instead of 1.5?0
     
  15. Jul 20, 2012 #14

    PhanthomJay

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    Yes.....1/2(74.5)(2.6)^2 + (74.5)(9.8)(2.5) = 1/2(74.5)v^2 + (74.5)(9.8)(0)

    Solve for v.
     
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