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Relatively simple mechanics problem

  1. Aug 7, 2015 #1
    1. The problem statement, all variables and given/known data

    A person pushes an object along an incline at an average speed of 15km/hr.After reaching the top, he pushes the object back from top to bottom, where the average speed is 30km/hr.if the length along which he pushes( hypotenuse of the inclined plane) is 4km and the force is constant, Find the height of the inclined plane.THE QUESTION CAN BE EASILY GRASPED BY LOOKING AT THE REFERENCE PIC.

    2. Relevant equations

    Don't know which one to use.
    3. The attempt at a solution
    Ok, so frankly i got nothing.but, since the force is constant and the speeds are different, i guess we can put F1=mv1/t1=mv12/s?
    And F2=F1+mgsinΘ? That's all i got.plz help
     

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    Last edited: Aug 7, 2015
  2. jcsd
  3. Aug 7, 2015 #2

    haruspex

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    Why do you think he has to push to get it back down?
     
  4. Aug 7, 2015 #3
    I know, he won't HAVE to push(the object would just slide,F=mgsinθ), but that's what the question says.BESIDES, since he applied the same force, v should have been same(in case of straight plane), but v2 is faster because(PROBABLY) F2=F1+mgsinθ
     
  5. Aug 7, 2015 #4

    haruspex

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    It doesn't say whether there is friction, so I considered there might be.
    Ignoring that for now, if he applies force F going up, what is the acceleration?
    If the distance to the top is s, what is the average speed?
     
  6. Aug 7, 2015 #5
    Neglect friction.sorry for not adding that.And,just for convenience, I converted the distance and time units(km and hr) to m and s.Now,average speed to the top is given.
    So, time=4000m/[(15*5/18)m/s]=960 seconds.
    ⇒a1=v/t=(15*5/18)/960=0.00434m/s2
    What next? do i find the mass(if so,how and why?)?
     
  7. Aug 7, 2015 #6

    haruspex

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    Please explain your calculation of the acceleration. What did you divide by what, exactly?
    Then write out the appropriate equation for the forces giving rise to this acceleration.
    It's late here, back tomorrow.
     
  8. Aug 7, 2015 #7
    LOL, sorry, didn't realize that(Australia-----diff. time zone*duh*)
    Is it wrong though????? I mean, s=v/t, where v=15km/hr=15*5/18 m/s, and s = 4000m, so t= s/v=960s(used the calc.).
    and, a=v/t=(15*5/18)/960=0.00434m/s^2.
     
  9. Aug 7, 2015 #8

    haruspex

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    I meant that last step. Where you have applied v/t, how would you define the velocity you used? What is the definition of v in a=v/t? What velocity does v represent?
     
  10. Aug 7, 2015 #9
    so,i have been trying something, is it possible that i could first find t1=960s,t2=480s, then the average speeds are 0.24m/s and 0.12m/s, and 4000F=mgh?
     
  11. Aug 7, 2015 #10

    haruspex

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    what other work has F done?
     
  12. Aug 8, 2015 #11
    don't know.....moved the object against the component of the gravitational force?????
     
  13. Aug 8, 2015 #12
    CORRECTION: the average speeds are------- 4000/960=4.1666....m/s and 4000/480=8.3333....m/s.
     
  14. Aug 8, 2015 #13

    haruspex

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    As you calculated, the object accelerates up the slope. Since the force is constant, that acceleration is constant.
    The question as stated could be clearer. You need to assume that the object starts from rest in each direction, and is somehow brought to a sudden stop at the end in each case. So, what other form of energy do you need to consider?
     
  15. Aug 9, 2015 #14
    Kinetic?
     
  16. Aug 9, 2015 #15

    haruspex

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    Yes.
     
  17. Aug 10, 2015 #16
    Are you implying that 4000F= mv2/2???? If so, which speed is v?
     
  18. Aug 10, 2015 #17

    haruspex

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    No, I'm saying that in both directions the object starts with no KE but finishes with KE.
    Going uphill, it also gains PE, but loses it downhill.
    You can work out the final speed in each case from the average speed and the fact that forces are constant.
     
  19. Aug 10, 2015 #18
    Ok final speed in first case will be twice of average speed( because initial speed is zero)=25/3m/s.so,KE1=mv2/2=625m/18.
    And final speed in second case will be 2×30×5/18=50/3m/s.
    Now what?the 2 KEs are different.I'm not sure, but it won't be W=KE2-KE1, right?because that's a different thing.
     
  20. Aug 10, 2015 #19

    haruspex

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    So correct your energy equation in post #9.
    The two phases, uphill and downhill, are completely separate. There is no carry over of KE from the first to the second. Why would you take the difference of the two KEs?
     
  21. Aug 10, 2015 #20
    But what should I do next?How do I relate these KEs with the height of the incline?
     
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