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Babynini

  • one year ago

Force and work.

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  1. Babynini
    • one year ago
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    A car drives at 800ft. on a road inclined at 9 degrees. Car weighs 2700lb. Thus gravity acts straight down on the car with a constant force of f=-2700j. Find the work done by the car in overcoming gravity.

  2. Babynini
    • one year ago
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    Um..what i've done so far is |v|=2700cos(9) = 2666.76 which is the force experienced by the driveway. Is that the answer i'm looking for?

  3. Babynini
    • one year ago
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    @ganeshie8

  4. Babynini
    • one year ago
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    @zepdrix :)

  5. Babynini
    • one year ago
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    @jim_thompson5910

  6. jim_thompson5910
    • one year ago
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    Work = Force * Displacement

  7. jim_thompson5910
    • one year ago
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    Gravity is pulling straight down, so when we think of "displacement" we are only considering the vertical displacement of the car

  8. jim_thompson5910
    • one year ago
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    |dw:1433399821064:dw|

  9. jim_thompson5910
    • one year ago
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    when you wrote `Thus gravity acts straight down on the car with a constant force of f=-2700j` that doesn't make any sense. Joules aren't a measure of force. They are measure of energy. You might be thinking of newtons?

  10. Babynini
    • one year ago
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    uhh idk that's what my homework says.

  11. jim_thompson5910
    • one year ago
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    oh nvm, -2700j is a vector, not 2700 joules

  12. Babynini
    • one year ago
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    haha yeah

  13. Babynini
    • one year ago
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    where is hte 800 from?

  14. jim_thompson5910
    • one year ago
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    "A car drives at 800ft"

  15. jim_thompson5910
    • one year ago
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    I'm assuming it drives up that incline and drives 800 ft

  16. Babynini
    • one year ago
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    ah yeah k.

  17. jim_thompson5910
    • one year ago
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    use trig to find the value of y in the drawing below |dw:1433400113262:dw|

  18. Babynini
    • one year ago
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    125.148 ?

  19. jim_thompson5910
    • one year ago
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    so the car essentially goes up vertically 125.148 ft this is the vertical displacement we care about

  20. Babynini
    • one year ago
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    (ps: it says you are expected to use vectors and the appropriate formulas to find work. Just as a note. :P)

  21. Babynini
    • one year ago
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    oo ok

  22. Babynini
    • one year ago
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    so now we have 2=-2700 * 125.148

  23. Babynini
    • one year ago
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    that 2 was meant to be a w

  24. Babynini
    • one year ago
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    er. that will make a very nasty negative number..

  25. jim_thompson5910
    • one year ago
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    were you given the vector formula for work in physics?

  26. Babynini
    • one year ago
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    This is precalc.

  27. Babynini
    • one year ago
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    give me a moment to find it, I have a formula somewhere :)

  28. Babynini
    • one year ago
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    Yep w=f*d

  29. jim_thompson5910
    • one year ago
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    I'm thinking you basically take the force needed to climb that vertical distance and multiply it with the vertical distance I'm not sure how to fit in vector formulas though

  30. Babynini
    • one year ago
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    so an example given is: A man pulls a wagon horizontally by exerting a force of 20 lb on the handle. If the handle makes an angle of 60 degrees with horizontal, find the work done in moving the wagon 100ft Solution: We choose a coordinate system with the origin at the initial position of the wagon. That is, the wagon moves from point p(0,0) to the point Q(100,0) the vector that represents this displacement is d=100i The force on the handle can be writen in terms of components as F= (20cos(60))i + (20sin(60))j F = 10i+10 sq3j Thus the work done is (10i+10 sq3j)*(100i)=1000 = 1000

  31. Babynini
    • one year ago
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    so our displacement is D = 800i

  32. Babynini
    • one year ago
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    so it's written as F=(2700cos9)i+(2700sin9)j = 2666.785i+422.373j

  33. Babynini
    • one year ago
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    and work done is = (2666.785i+422.373j)*(800i)

  34. Babynini
    • one year ago
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    I'm not sure how they calculated the last part though.

  35. Babynini
    • one year ago
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    Is this kind of making sense? I'm following formulas mostly. o.0

  36. Babynini
    • one year ago
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    @jim_thompson5910 :P

  37. Babynini
    • one year ago
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    (sorry, i'll be patient ;))

  38. jim_thompson5910
    • one year ago
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    sorry I got distracted for a sec but it looks like you're on the right track. You dot product the force vector F and the displacement vector d to get the work

  39. Babynini
    • one year ago
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    it's ok! I know you're helping other people. Keep up the hard work :)

  40. Babynini
    • one year ago
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    so is 2133829.189 the final answer you're getting?

  41. jim_thompson5910
    • one year ago
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    The force of gravity is F = 0i + (-2700j) essentially this vector pulls the car straight down with magnitude 2700 lb -------------------- The displacement vector d = 800*cos(9)i + 800*sin(9)*j -------------------- dot product the two vectors W = F dot d W = [0i + (-2700j)] dot [800*cos(9)i + 800*sin(9)*j] W = [0*800*cos(9)] + [-2700*800*sin(9)] W = -337,898.444486899 That seems like a pretty massive number, so I'm not 100% sure

  42. Babynini
    • one year ago
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    bleh yeah that's pretty huge :/

  43. Babynini
    • one year ago
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    wait why 800 instead of 2700 before the cos and sin?

  44. jim_thompson5910
    • one year ago
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    because if we ignore the weight, and we just consider moving along the hypotenuse, we basically have this triangle |dw:1433402120340:dw|

  45. Babynini
    • one year ago
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    but in the example one they used the weight of the cart in that spot of the equation, which is why I ask. hm but yeah I see what you're saying.

  46. jim_thompson5910
    • one year ago
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    |dw:1433402181444:dw|

  47. jim_thompson5910
    • one year ago
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    |dw:1433402210524:dw|

  48. jim_thompson5910
    • one year ago
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    hmm now I'm not sure. I've looked back at your example and you're right

  49. Babynini
    • one year ago
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    lol. What is this madness.

  50. anonymous
    • one year ago
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    \[W=\vec F. \vec s=Fs \cos(\theta)\]

  51. anonymous
    • one year ago
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    Everything is given, just convert to standard units and use formula, feet to metres, pounds to kilograms and degrees to radians

  52. anonymous
    • one year ago
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    |dw:1433411117463:dw|

  53. Babynini
    • one year ago
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    @rational how do I solve this? There's so many formulas that have been given here D:

  54. Babynini
    • one year ago
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    F = -2700j so = <0,-2700> Displacement (or shift) = 800 feet. so = <800,0> all correct so far?

  55. jim_thompson5910
    • one year ago
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    <800,0> means you go 800 ft to the right and 0 ft up/down

  56. jim_thompson5910
    • one year ago
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    <800,0> = 800i + 0j

  57. Babynini
    • one year ago
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    Yep yep. So now according to W = F * D I dot product <0,-2700> * <800,0> ?

  58. jim_thompson5910
    • one year ago
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    but the object isn't going on a flat level plane its on an inclined plane

  59. Babynini
    • one year ago
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    oh yeh.

  60. Babynini
    • one year ago
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    where is that incorporated?

  61. jim_thompson5910
    • one year ago
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    think of the point \(\Large (r, \theta) = (800, 9^{\circ})\) what is the rectangular (x,y) form of this point?

  62. Babynini
    • one year ago
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    what?

  63. jim_thompson5910
    • one year ago
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    |dw:1433463727512:dw| what are x and y?

  64. Babynini
    • one year ago
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    Opposite and Adjacent? is that what you're asking?

  65. jim_thompson5910
    • one year ago
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    Once you find x and y, you will be able to represent the displacement as a vector <x,y>

  66. jim_thompson5910
    • one year ago
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    you use trig to find x and y sin(theta) = opp/hyp sin(9) = y/800 y = 800*sin(9) do the same with x (but use cosine instead)

  67. Babynini
    • one year ago
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    y= 125.149 x=790.151

  68. jim_thompson5910
    • one year ago
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    |dw:1433464222924:dw|

  69. jim_thompson5910
    • one year ago
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    so the displacement vector is <790.151, 125.149> or 790.151i + 125.149j

  70. Babynini
    • one year ago
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    and Force still is = <0,-2700> ?

  71. jim_thompson5910
    • one year ago
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    I'm realizing now that's where the mistake is that's the force of gravity, but we want the force of the car if it asked "what is the work done by gravity?" then we would use F = <0,-2700>

  72. jim_thompson5910
    • one year ago
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    but instead it says "Find the work done by the car in overcoming gravity."

  73. Babynini
    • one year ago
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    aah hrm. so something to do with the weight?

  74. Babynini
    • one year ago
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    Wait, but it says "the gracity acts straight down on the car with a constant force of -2700j." the force pushing against the car would be the same the car is pushing back though. sot hat would be the force the gar is "pushing against gravity"?

  75. Babynini
    • one year ago
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    *gravity

  76. jim_thompson5910
    • one year ago
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    maybe to overcome the force of -2700j, the car needs to apply 2700 lb somehow?

  77. jim_thompson5910
    • one year ago
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    yeah as you just said

  78. jim_thompson5910
    • one year ago
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    I'm not sure though

  79. Babynini
    • one year ago
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    But -2700 is what is being pushed against the car. Which is the same amount the car is pushing back against gravity. So that should still be the force.

  80. jim_thompson5910
    • one year ago
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    that might be correct

  81. Babynini
    • one year ago
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    ...

  82. jim_thompson5910
    • one year ago
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    sorry I'm not that great at physics

  83. Babynini
    • one year ago
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    apparently neither am I ;P

  84. jim_thompson5910
    • one year ago
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    if you use that link's formula, be careful to make sure things are converted to metric

  85. Babynini
    • one year ago
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    er but this is vector stuff not joules and what not

  86. jim_thompson5910
    • one year ago
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    pounds ---> kilograms feet ----> meters

  87. jim_thompson5910
    • one year ago
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    joules is just the units of work in physics

  88. Babynini
    • one year ago
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    Ah k.

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