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calculusxy

  • one year ago

MEDAL!!! How much work is done by an elevator that carries a person with a mass of 70 kg up 40 meters? Assume the elevator has a constant acceleration of 2 m/s^2.

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  1. calculusxy
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    The elevator is not a inertial system, nevertheless, we can consider it, as an inertial system if we introduce the fictitious force -m*Ac where Ac is the ecceleration of the elevator, and m is the mass of the person inside the elevator. So the situation is like below: |dw:1437833008631:dw|

  3. Michele_Laino
    • one year ago
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    in other word, on the person are acting two forces: the weight force whose magnitude is m*g= 70*9.81=...newtons the fictitious force, whose magnitude is m*Ac=70*2=...newtons Since the elevator is a inertial system, the laws of physics apply unchanged in it, so the requested work is: \[\Large W = m\left( {g + {A_c}} \right)h = ...joules\] where h=40 meters

  4. calculusxy
    • one year ago
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    So we are learning about energy now and the formulas that we have gone through so far are: Power = w/t Work = f x d Potential Energy: mass x gravity x distance Kinetic Energy: 1/2mv^2

  5. calculusxy
    • one year ago
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    The teacher said that he incorporated these formulas into the questions, so i need to understand what formulas go to what question to get the answer.

  6. Michele_Laino
    • one year ago
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    you can apply this formula: Work = f x d providing that the force f is: f= m(Ac+g) where Ac=2 m/sec^2

  7. calculusxy
    • one year ago
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    Why can't we use the formulas to find the power because that this the amount of work done in an interval?

  8. calculusxy
    • one year ago
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    *formula

  9. Michele_Laino
    • one year ago
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    the elevator is not an inertial system, since there are 2 forces acting on the person inside the elevator. I think it is better if we use the formula: work = force * distance

  10. calculusxy
    • one year ago
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    I still don't understand how you find the force.

  11. Michele_Laino
    • one year ago
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    In other words, I have applied the Principle of Equivalence, and I consider the elevator like an inertial system, providing that to introduce the fictitious force

  12. calculusxy
    • one year ago
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    i am sorry , but i don't know the principle of equivalence because i am just a rising eighth grader.

  13. Michele_Laino
    • one year ago
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    when you use an elevator in order to go up, at the starting of motion, you feel like pressed towards the floor of the elevator, right?

  14. calculusxy
    • one year ago
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    yes

  15. Michele_Laino
    • one year ago
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    that is due to the fictitious force which is acting on you

  16. calculusxy
    • one year ago
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    ok

  17. Michele_Laino
    • one year ago
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    that fictitious force will add to your weight force

  18. calculusxy
    • one year ago
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    ok

  19. calculusxy
    • one year ago
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    can we use the formula f = m x a?

  20. Michele_Laino
    • one year ago
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    so the total force which is acting on you is: fictitious force+weight force

  21. Michele_Laino
    • one year ago
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    and the work done by the elevator is the work done by that total force

  22. calculusxy
    • one year ago
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    \[Force = mass \times acceleration\] \[Force = 70 kg \times 2m/s^2\] \[Force = 140 N \] Work: \[Force \times Distance \] \[140N \times 40 m\] \[1600J\]

  23. Michele_Laino
    • one year ago
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    yes! the fictitious force is given by the subsequent formula: f=m*Ac where Ac=2 m/sec^2

  24. Michele_Laino
    • one year ago
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    now you have to add the work done by the weight force

  25. calculusxy
    • one year ago
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    Oh... i am sorry about misinterpreting it because i didn't know that Ac actually meant acceleration.

  26. Michele_Laino
    • one year ago
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    ok!

  27. Michele_Laino
    • one year ago
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    the work done by the weight force is: 70*9.8*40=...?

  28. calculusxy
    • one year ago
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    that's the potential energy right?

  29. Michele_Laino
    • one year ago
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    yes!

  30. calculusxy
    • one year ago
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    27440

  31. Michele_Laino
    • one year ago
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    the work done against the weight force is equal to the potential energy gained by the elevator

  32. Michele_Laino
    • one year ago
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    correct!

  33. calculusxy
    • one year ago
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    And now I have to add it: 27440 + 1600 = 29040

  34. Michele_Laino
    • one year ago
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    hint: 70*2*40=5600

  35. calculusxy
    • one year ago
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    i got lost after finding the force.

  36. Michele_Laino
    • one year ago
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    so, we have: 27440+5600=...

  37. calculusxy
    • one year ago
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    but why did we need to do 70*2*40?

  38. Michele_Laino
    • one year ago
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    the work done by the fictitious force, is : m*Ac*h= 70*2*40=...

  39. Michele_Laino
    • one year ago
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    \[{\text{work = force }} \times {\text{ distance = }}\left( {{\text{m}} \times {{\text{A}}_{\text{c}}}} \right) \times {\text{h}}\]

  40. calculusxy
    • one year ago
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    give me a minute...

  41. calculusxy
    • one year ago
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    so i do 27440 +5600= 33040

  42. Michele_Laino
    • one year ago
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    that's right!

  43. calculusxy
    • one year ago
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    so that's my answer right?

  44. Michele_Laino
    • one year ago
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    yes!

  45. calculusxy
    • one year ago
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    thank you!!! i have another question may i ask that as well?

  46. Michele_Laino
    • one year ago
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    ok!

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