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Work Done by serveral forces

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance of 20m along level ground. The total weight of sled and load is 14700N. The tractor exerts a constant 5000N at an angle of 36.9 above the horizontal. There is a 3500N friction force opposing the sled's motion. Find the work done by each force acting on the sled and the total work done by all the forces.

    3. The attempt at a solution

    To start off,

    I found the work done by each force.

    I drew the figure as follows:


    my calculations

    cos(angle) = x/5000

    5000 cos 36.9 = 3998.42

    (3998.42J)(20m) = 79968.47


    3500cos180 = -3500

    (-3500J) (20m) = -70000

    79968.47-70000 = 9968.47

    My answer: 9968.47
    Book answer: 10000

    Is my process wrong, if so where did I go wrong? Don't the gravity and normal force cancel out?

    From my reading in the book, we suppose to only count the horizontal forces, which is what I did here.
  2. jcsd
  3. Oct 22, 2007 #2


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    Gold Member

    I agree with your answer, what is the answer in the book?
  4. Oct 22, 2007 #3
    the answer in the book is 10,000.

    Don't know if this is rounded off,etc...since physic books usually tend to teach that we must not round off and be exact upon things.
  5. Oct 22, 2007 #4
    You are probably correct. My book actually does strange things such as G=9.8 instead of G=9.81. I'm pretty used to my answers being slightly off and I'm given full credit along with my classmates.
  6. Oct 22, 2007 #5


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    Gold Member

    Your answer is more than close enough. The book is taking into account significant figures.
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