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Work Done

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the work done by the two forces together in moving an object a distance of d = 9 m as shown in the diagram? The magnitude of each force is 21 N.
    There is a picture that comes with this
    http://capa-new.colorado.edu/teacher/capalibrary/Graphics/Gtype12/prob39.gif

    2. Relevant equations
    So i know that W==Fnet*d but i am unsure of how to do it with two angles. i also know that work can be found by W=Fcos(theta)d and i have tried both and neither seem to work.


    I also have a second question that is similar
    With brakes fully applied, a 1430 kg car deccelerates from a speed of 83.0 km/hr. What is the work done by the braking force in bringing the car to a stop?

    So the way i tried to find this was by doing W=mad and that came up as wrong.

    and a second part to that question is
    What is the change in kinetic energy of the car

    I know that KEI+PEI=KEF+PEF
    so would i just do 1.2mvsquared to find the KEF?
     
    Last edited: Oct 20, 2008
  2. jcsd
  3. Oct 20, 2008 #2

    mgb_phys

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    The forces simply add.
    Calculate each force in the direction of 'd' ( this is equvialent to the length of F in that direction) then add them for the total force.
     
  4. Oct 20, 2008 #3
    so for the first question i should do
    (21cos(45)*9)+(21cos(45)*9)?
     
  5. Oct 20, 2008 #4
    oh nm about the second question i figured out what was wrong. i forgot to make it in m/s and kept it in km/hr. my bad
     
  6. Oct 20, 2008 #5
    please can someone help me!
     
  7. Oct 20, 2008 #6

    mgb_phys

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    Isn't the change in KE just the initial ke of the car, since it has no ke at the end.
     
  8. Oct 20, 2008 #7
    i figured out that second part already, it was the first part that was giving me trouble. the part about what is the work done by the two forces together. i mean would it be 0J since it is technically a 90deg angle?
     
  9. Oct 21, 2008 #8
    ok so i was told to do W=Fdcos(theta) but what i am not sure of is do i basically multiply by 2 cause there are two angles or do i just do the one?
     
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