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Work, distance and time problem

  1. Dec 15, 2011 #1
    1. A 133 watt motor is used to lift a 148 kg container to a height of 6 m. How much time does it take to lift the container with the motor?




    2. work= Fd; P= work/t; F=ma



    3. I am thinking way too much into this. I converted the watts into J/s,[1W=1J/s] = 133 J/s. I converted the weight into mass [mass = weight / gravity] = 15.1 kg. Now I am stuck because I need to know the force, [F = ma] but dont have the information for acceleration. I totally confused myself and dont know what to do. I believe I need to divide the work by the power to solve for time, but I am stuck on figuring out the work
     
  2. jcsd
  3. Dec 15, 2011 #2
    Work is force times distance. You know the force, you know the distance.
     
  4. Dec 15, 2011 #3
    Just figured that out. I told you I was reading too far into this, lol.
     
  5. Dec 15, 2011 #4
    So what I have now is: time = work / power; t = 888J / 133J/s leaving 6.6766...s (the joules cancel out) which we round to one decimal place for this question, and my answer would be 6.7 seconds. Is this correct?
     
  6. Dec 15, 2011 #5
    You are off by a factor of g. You already have the mass of 148 kg. Why are you dividing by g?
     
  7. Dec 15, 2011 #6
    The work is not 888J.
     
  8. Dec 15, 2011 #7
    148kg is not the mass, it is the weight. I forgot that this weight (148) is also the force in newtons. I did not need to convert it to mass, it was already in the form I needed it. [F = W = mg]. If all I had was the mass, I would have multiplied the mass times gravity to obtain the Force, or weight.
     
  9. Dec 15, 2011 #8
    A kilogram is a unit of mass, not weight.
     
  10. Dec 15, 2011 #9
    Good point, force is either pounds or newtons. Thanks for pointing that out.
     
  11. Dec 15, 2011 #10
    Reworking the problem:
    F= 1450.4N (148*9.8)
    W= 8702.4J (1450.4N*6m)
    T= 65.4s (8702.4J/133J/s)

    I think I am still missing something.
     
  12. Dec 15, 2011 #11
    Like what? Power is the rate at which work is done. You calculated how much work there is to do. What you did in your last post is correct.
     
  13. Dec 15, 2011 #12
    Woohoo, thank you for your guidance! I always want to understand the work I am doing and it kills me when I cant figure something out. I have spent 2 hours on this one problem. I really appreciate you pointing out my errors and explaining why. Thats something you cant get when taking an online class.
     
  14. Dec 15, 2011 #13
    Good fortune to you in your endeavor. Bye.
     
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