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calculusxy

  • one year ago

Let's say Jack is running in two 5k races (for the uninitiated, 5k races are 5km long). And in the one, Jack carries a water bottle weighing 2N and he finishes in 32 min. IN the second one, he carries the same water bottle, but finishes in 30min. How much work did Jack do by carrying that water bottle in race 1? Race 2? How much power did he exert in race 1? Race 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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    I think that if the velocity is constant, along each race, then the work done is null, in both races

  3. Michele_Laino
    • one year ago
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    the gravity is perpendicular with respect to the direction of the race

  4. Michele_Laino
    • one year ago
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    |dw:1437835911067:dw|

  5. calculusxy
    • one year ago
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    can we somehow formulate the work using force x distance ?

  6. Michele_Laino
    • one year ago
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    if distance and force are mutually perpendicular, then the work done by that force is null

  7. Michele_Laino
    • one year ago
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    so, the work done by the weight force of the bottle is null, then also the work done by the Jackis null

  8. Michele_Laino
    • one year ago
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    oops...Jack is...

  9. calculusxy
    • one year ago
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    i don't understand how the force is null if it is perpendicular.

  10. Michele_Laino
    • one year ago
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    the weight force is not null what is null is the component of the weight force along the direction of the race

  11. anonymous
    • one year ago
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    If he starts at rest, then he does work on the water bottle to speed it up. If he ends at rest, then he does work on the water bottle to slow it down. The work cancels out to zero.

  12. anonymous
    • one year ago
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    From an energy perspective, we know the energy of the bottle doesn't change, so the net work would have to be zero.

  13. calculusxy
    • one year ago
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    so i tried to work with 2N and 5,000m = 10000J for both. how do i show it to be canceling out?

  14. anonymous
    • one year ago
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    Weight it perpendicular to the direction he ran.

  15. Michele_Laino
    • one year ago
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    please, note that the writing 2*5000= 10,000 has no meaning, since the weight force and the direction of the race are mutually nperpendicular

  16. calculusxy
    • one year ago
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    okay

  17. anonymous
    • one year ago
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    You'd want to do \(2~\text{N}\cdot 5~\text{km} \cdot \cos(90^\circ )= 0~ \text{J}\).

  18. calculusxy
    • one year ago
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    we didn't learn trig yet

  19. calculusxy
    • one year ago
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    what about for the power?

  20. Michele_Laino
    • one year ago
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    it is suffice that you keep in mind that when force and distance are mutually perpendicular, then the work done by that force is zero

  21. calculusxy
    • one year ago
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    so for power can we do 10000J/32 min = 312.5 watts

  22. Michele_Laino
    • one year ago
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    the power also is null, since the work is null

  23. Michele_Laino
    • one year ago
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    remember that: power= work/ time

  24. calculusxy
    • one year ago
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    oh yes

  25. calculusxy
    • one year ago
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    thank you so much! i have a last question and then i am done for the day

  26. Michele_Laino
    • one year ago
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    ok!

  27. calculusxy
    • one year ago
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    a new conveyor system at the local packaging plant will use a motor-powered mechanical arm to exert an average force of 1000N to push large crates a distance of 10 meters in 30 seconds. determine the power output required of such a motor.

  28. Michele_Laino
    • one year ago
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    here the average speed is: v=10/30=...m/sec

  29. calculusxy
    • one year ago
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    1m/sec?

  30. Michele_Laino
    • one year ago
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    hint: 10/30= 1/3 =...?

  31. Michele_Laino
    • one year ago
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    \[10:30 \cong 0.334m/\sec \]

  32. Michele_Laino
    • one year ago
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    now the requasted power is: \[{\text{P = Force }} \times {\text{ speed = }}1000 \times 0.334 = ...watt\]

  33. Michele_Laino
    • one year ago
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    requested*

  34. calculusxy
    • one year ago
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    334 watts?

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

  36. calculusxy
    • one year ago
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    so my answer is 334 watts right?

  37. Michele_Laino
    • one year ago
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    yes! more precisely: P= 333.34 watts

  38. calculusxy
    • one year ago
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    THANK YOU TREMENDOUSLY!!!!!!!!!!!!!!!!

  39. Michele_Laino
    • one year ago
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    :) :)

  40. calculusxy
    • one year ago
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    Ciao

  41. Michele_Laino
    • one year ago
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    Ciao! Hi!

  42. calculusxy
    • one year ago
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    how do you say "bye"

  43. calculusxy
    • one year ago
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    addio?

  44. Michele_Laino
    • one year ago
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    yes! I say "Ciao"

  45. calculusxy
    • one year ago
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    oh okay. thank you! learned a little bit of italian as well. couldn't get any better !!

  46. Michele_Laino
    • one year ago
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    ok! :)

  47. calculusxy
    • one year ago
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    A presto

  48. Michele_Laino
    • one year ago
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    see you soon!!

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