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Homework Help: Need help on finding the work done on an Object by the applied force.

  1. Jun 12, 2012 #1
    "A 50 kg crate is being dragged across a floor by a force of 225 N at an angle of 40° from the horizontal. The crate is dragged a distance of 5.0 m and the frictional force is 60 N."

    Now from that information i am supposed to find the work done by the applied force on the object.

    This is what i have so far. Since W= (F cos θ)Δd, i did W= (225 N cos 40°)5.0 m ≈ 862 N.m
    Am i correct

    PS this is my first time on physics forum. :)
  2. jcsd
  3. Jun 13, 2012 #2


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    welcome to pf!

    hi xuhhad! welcome to pf! :smile:
    yes :smile:

    (what is worrying you about that?)
  4. Jun 13, 2012 #3

    Simon Bridge

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    yep - hi xuhhad! welcome to PF!
    It can be a bit daunting at first - especially when an answer seems to come too easily. As you learn more you'll be able to develop strategies to figure out if you are right or not without having to ask anyone.

    I suppose it next asks for the work done by the frictional force?
  5. Jun 13, 2012 #4
    Yes it does :/

    and again i am stuck sorry for the inconvenience. I really need help.
    Last edited: Jun 13, 2012
  6. Jun 14, 2012 #5

    Simon Bridge

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    Funny that :) You have to think it through - what is the definition of work in terms of force? Can you write it in words?
  7. Jun 14, 2012 #6


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    hi xuhhad! :smile:

    you seem to have a lack of confidence

    but your original answer, using the correct definition of work done, was fine, so just do the friction in exactly the same way, and then that'll be fine too! :biggrin:

    what do you get? :smile:
  8. Jun 14, 2012 #7
    hmm the work formula would be W=mg/Δd
  9. Jun 15, 2012 #8

    Simon Bridge

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    Get used to putting the formulas into words and thinking about what they mean.

    Your formula has work equal to the vertical force divided by the displacement ... which would mean that the further you drag the crate, the less energy is taken by friction ... does that sound right?

    Aren't you also lifting the crate a bit as you pull it? Does that matter?

    (Since you have not been given dimensions for the crate, I expect that you are to treat all forces through the center of mass?)
  10. Jun 15, 2012 #9


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    no, work done = force "dot" displacement

    (so it can be positive or negative)

    learn this definition!! :smile:
  11. Jun 17, 2012 #10
    i still don't understand how i can find the frictional force.
  12. Jun 17, 2012 #11


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    ?? :confused:

    the frictional force is given in the question …

    it's 60 N horizontally, "backwards"​
  13. Jun 18, 2012 #12
    i get it i just have to use the work formula again, something like this correct

    W=(-Ff)cos θ * Δd

    Thak u for all the help :approve:
  14. Jun 18, 2012 #13
    F_kinetic = μ_k F_normal

    Does that help?
  15. Jun 18, 2012 #14

    This part of your question sounds like:

    The problem
    A=1, B=2, AB

    Relevant equations

    Am I correct?

    P.S. I'm not hijacking this thread. I obviously know what 1 multiplied by 2 is. I'm just saying that the problem I wrote is very similar to the problem given by the OP.
  16. Jun 18, 2012 #15
    Yes thank u for posting i already completed the question thank u for the help though
  17. Jun 18, 2012 #16
  18. Aug 4, 2012 #17
    if the is a force acting opposite to the direction of motion of the crate ( frictional force= 60N) this implies that the total force pulling the crate is now; 225N - 60N =165N.
    the reaction force to the floor will be= 50cos40=38.30N
    Therefore, total pulling force= 165N - 38.30N= 126.7N.
    Now Work done= 126.7 x 5 = 633.5J
    Last edited: Aug 4, 2012
  19. Aug 4, 2012 #18
    Pls i hope you have seen where your errors are coming from. You forgot about the reaction force onthe crate and also the frictional force opposing the force causing the motion of the crate. Thanks
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