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Homework Help: Calculate work done

  1. Jan 16, 2010 #1
    1. A 40kg stone is pulled slowly up a smooth ramp inclined at 30o to the horizontal. It stops at the top, at a height of 100m above its initial height. Calculate the work done by:
    a) Using 'work done = force x distance moved in direction of force'.
    b) Using conservation of energy




    2. F=ma
    F-Fgravity=mgsin(theta)
    Work done=Fscos(theta)




    3. Weight = 400N
    F-400=400sin(30)=600N
    Work done = 600 x 100cos(30) = 52kJ (2sf)


    Not sure this is correct
    please advise?
     
  2. jcsd
  3. Jan 16, 2010 #2

    tiny-tim

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    Hi lemon! :smile:
    No … where did the " - 400 " come from?
     
  4. Jan 16, 2010 #3
    weight - mg = 40kg x 10m/s
     
  5. Jan 16, 2010 #4
    Right. Sorry. To calculate the Fnet.
    Fnet=400sin(30) = 200N
    Work done = 200N x 100m = 20kJ
     
  6. Jan 16, 2010 #5

    tiny-tim

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    F yes :smile:, W no :redface:.

    'work done = force x distance moved in direction of force'

    so the "x distance" isn't 100.
     
  7. Jan 17, 2010 #6
    ahh!
    sin(theta)=opposite/hypotenuse
    100/sin(30) = 200m

    W=200N x 200m =4000J
    or 4kJ
     
  8. Jan 17, 2010 #7

    tiny-tim

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    That's it!

    btw, you might like to try this, just to see how it all fits together …

    you've assumed that the force is applied the obvious way, by a rope parallel to the slope,

    but suppose it is applied by a rope at an angle φ to the slope (φ + 30º to the horizontal) …

    what is F then, and what is W? :smile:

    And of course try b) (calculate the work done using conservation of energy)
     
  9. Jan 17, 2010 #8
    So for b) calculate the work done using conservation of energy

    Total amount of energy always stays the same, as energy cannot be created nor destroyed, just transfered to other forms.

    We have gained height by inputting energy to pull the block up the slope.
    So we have gone from kinetic energy to gravitational potential energy. We have calculated that the work done is 4kJ. So:
    KE lost = PE gained
    4kJ = mgh

    I think I'm lost :uhh:
     
  10. Jan 17, 2010 #9

    tiny-tim

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    You can assume the KE is zero all the time …
    So the work done is just the difference in PE, which is … ? :wink:
     
  11. Jan 17, 2010 #10
    mgh - 4kJ
    (40kg x 10m/s x 100m) - 4kJ
    = 36000j
    36kJ
     
  12. Jan 17, 2010 #11

    tiny-tim

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    No, the work done is just the difference in PE

    (oh, and I've just noticed you miscounted the 0s in …)
     
  13. Jan 17, 2010 #12
    How can there be a difference in PE when there is no PE at the bottom of the slope?
     
  14. Jan 17, 2010 #13

    tiny-tim

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    There isn't no PE … the PE is 0.

    So the difference is … ? :smile:
     
  15. Jan 17, 2010 #14
    40kJ - 0?
     
  16. Jan 17, 2010 #15

    tiny-tim

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    yes, the work done is just mgh.

    "Work done" is another name for the difference in PE :wink: (in a conservative field).

    ok, your 40kJ is now the same as you should have got under part a). :smile:
     
  17. Jan 17, 2010 #16
    Man! These questions are brain twisters. I look at them and think they look straight forward, but man!! I'm really glad I decided to come on here and get help with these or I don't think I would hardly have understood at all. Is it just me?

    Last part:
    It is now dropped from the top of the building descending vertically in free fall. Calculate its speed just before it hits the ground. What happens to its kinetic energy?

    This is PE to KE
    mgh = 1/2mv2
    40kJ = 1/2x40xv2
    root2000 = 20root5 or 44.7m/s

    Kinetic energy is then transfered into any deformation of the ground/stone, and into sound and heat. Is that what the question is looking for?
     
  18. Jan 17, 2010 #17

    tiny-tim

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    Yes, that's fine. :smile:
     
  19. Jan 17, 2010 #18
    Thanks again tT
     
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