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Spring. (Potential energy and work done)

  1. Mar 24, 2013 #1
    Please see attached images.

    Those pencil-written answers are directly copied from marking scheme (after using considerable amount of time calculating and still getting the wrong answer)

    1. For question 1, why is the weight taken as 3.8N regardless of the additional force F?

    2. In question 1, gravitational potential energy is supposed to be lost because the height of the object from the ground has decreased. Then why is the answer not negative?

    3. For question 3, why is it just Es - Ep? Combining with my second question, is it because the F caused a decrease in gravitational potential energy, and caused an increase in elastic potential energy at the same time?
     

    Attached Files:

  2. jcsd
  3. Mar 24, 2013 #2
    Strictly speaking, Δh is defined as [itex]h_f - h_i[/itex] and hence in this case it is negative thus making ΔE negative.
    Net work done by the force is the net change in its mechanical energy in this case and hence it is [itex]ΔE_s + ΔE_p[/itex] .
     
  4. Mar 24, 2013 #3

    TSny

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    1. "Weight" is the force of gravity acting on the mass; i.e., the force which the earth pulls down on the mass.

    2. You're right, the change in gravitational PE is negative.

    3. That's correct.
     
  5. Mar 25, 2013 #4
    So in the third diagram, the spring actually has MORE total energy compared to the previous two diagrams?
     
  6. Mar 25, 2013 #5
    Right
     
  7. Mar 25, 2013 #6

    TSny

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    Yes, the potential energy stored in the spring is greatest in the third diagram. For a spring, the more it's stretched the more energy it stores.
     
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