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Yr 11 physics HW

  1. Sep 8, 2008 #1
    Hey i need some help with this question please:

    A car of mass 2.135 * 10^3 travels at a constant speed up a slope of 30 degrees a distance of 25.3m. If the frictional resistance is 0.100 of the weight of the car, find:
    (a) the work done
    (b) the energy expended

    The answer is 3.18 * 10^5 J

    The formulas we have been using are w=fs and f=ma also si units please.
  2. jcsd
  3. Sep 8, 2008 #2


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    Homework Helper

    What is the weight of the car?
    What is the frictional resistance?
    what is the component of the weight of the car along the slope?
    What is the total force?
  4. Sep 8, 2008 #3
    i can figure the weight out using f=ma but i keep getiing the wrong answer no matter what i do??
  5. Sep 8, 2008 #4
    here' s what ive done so far.
    weight = 2.135*10^4 N
    frictional resistance = 2.094*10^3
    and i try to put this into w=fs cos theata
    and i get 4.589*10^5

    please help
  6. Sep 8, 2008 #5


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    Weight = 2.135*10^3*9.8 = 20923 N
    The component of the weight of the car along the slope = 20923 N*sin(30 degree)
    Total force = 20923 N*sin(30 degree) + 20923*0.100 N
    Now find the work done.
  7. Sep 9, 2008 #6
    hey thanks alot for the help what i missed was that when you calculate the workdone it is the total force and not just the weight. thanks again
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