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Why is the work in this question = to the potential energy?

  1. Nov 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass of 235kg is lifted by a winch and an electric motor. The motor does 1800J of work on the mass to raise it to an upper floor. The electric energy needed to complete this task is 160,000J

    2. Relevant equations
    I know the solution but I don't why it is the solution
    Solution:
    m*g*h = W
    h = 1800J/(235kg*9.5N/Kg)
    = 0.78m

    3. The attempt at a solution
    I think it's because the mechanical energy (Em= Ek + Eg) is 1800J and when the mass is at the top all the energy is in the form of gravitational potential energy because the velocity is 0, making the kinetic energy 0.
    I want to know why so if on a test there is a similar question I know how to do it
     
  2. jcsd
  3. Nov 25, 2015 #2

    andrewkirk

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    That's a reasonable enough answer.
     
  4. Nov 25, 2015 #3
    Does the problem statement ask you to find the height, and you want to know why you use the potential energy and not the electric energy?

    Why is the electric energy greater than the potential energy?
     
  5. Nov 25, 2015 #4
    Yeah I want to know why the work is used and not the electric energy. I'm pretty sure the the electric energy is more than the work because energy might be lost due to heat, friction, sound etc. The next part of the question asks you to find the efficiency of the system for which you divide the output energy (the work) by the input ( the electrical energy) and it comes out to be 1.1% or so.
     
  6. Nov 25, 2015 #5
    You're right! So the part that's lost doesn't do work lifting the mass.
     
  7. Nov 25, 2015 #6

    haruspex

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    There are two things that seem strange with the provided solution.
    0.78m is nowhere near enough to be the distance to a higher floor.
    160,000J spent to achieve 1800J of useful work is unbelievably inefficient.
    I note it says the motor does 1800J of work on the mass. As I understand it, the work done on a rigid body is the increase in KE of the body. The work done against gravity is work done on the Earth-mass system. Yes, presumably the mass eventually does come to rest, but that would be a result of applying a brake when the mass is high enough. So, arguably, the 1800J was work done in providing KE (ultimately thrown away by applying the brake) and the rest of the 160,000J was the work done against gravity. That at least gives a more reasonable answer, about 69m, 20 storeys say.

    Btw, 9.5m/s2 is rather an unusual value for g. It's generally taken as 10, 9.8 or 9.81.
     
  8. Nov 26, 2015 #7
    The 9.5 was just a typo. He used 9.8 in his calculation.

    While your scenario does sound more plausible, I note that he said he knew that the right answer was 0.78 m.

    It's possible that his teacher made an error transcribing the problem, that the original value was 18000 J, giving a constant-speed lift height of 7.8 m and an efficiency of 11%. Still pretty low, but not ridiculously so!
     
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