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Work done = 1 N x 1 m = 2 kg 1m2 1.414s-2 = 1 J ??? wrong ??

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  1. Jan 7, 2016 #1
    • Poster has been reminded to post schoolwork-type threads in the Homework Help forums
    ~In the vacuum of outer space~

    If a rocket motor exerts a force of 1 Newton on a mass of 2 kilograms for 1.414213562 seconds over a distance of 1 meter, how many Joules of work have been done?

    I thought
    Work = 1 N x 1 m = 1 kg 1m2 1s-2 = 1 J

    so
    Work = 1 N x 1 m = 2 kg 1m2 1.414s-2 = 1 J ???

    I think there is something wrong with my set-up, but what is the proper way to write it?
    Is my textbook wrong?
    Please see the attached .PNG file for further clarification. physics question.png
     
  2. jcsd
  3. Jan 7, 2016 #2

    A.T.

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    "1N per second" is nonsense.
     
  4. Jan 7, 2016 #3
    Ok, I think I understand.
    Let me re-word my question:

    After pushing 2 kilograms a distance of 1 meter, how many Joules of work has a 1 Newton motor (2 kg · 0.5 m/s2) done?

    -or-

    After running for 1.414213562 seconds, how many Joules of work has a 1 Newton motor (2 kg · 0.5 m/s2) done?

    Are these equivalent questions?

    And of course, how many Joules?
    (& please show the set-up!)
     
  5. Jan 7, 2016 #4
    And your data is inconsistent. If you accelerate the 2 kg with a force of 1N, you can do it along a distance of 1m OR during 1.4 s.
     
  6. Jan 7, 2016 #5

    russ_watters

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    Yes. Work is force times distance. In both cases, you have 1 N and 1m = 1J.
     
  7. Jan 7, 2016 #6

    russ_watters

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    Am I missing something? Doesn't applying a force of 1N cause a 2kg mass to travel 1m in 1.4s?
     
  8. Jan 7, 2016 #7

    HallsofIvy

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    What you are "missing" is that force causes acceleration not speed. A force of 1 N will cause a 2 kg mass to accelerate at 1/2 m/s^2. How far it will move in that time depends upon the initial speed. Perhaps you are assuming the initial speed was 0 but that was not given.
     
  9. Jan 7, 2016 #8
    Maybe I am missing something.
    The acceleration will be ## \frac{1N}{2kg} = \frac{1}{2} \frac{m}{s^2}## , right?
    The distance traveled:
    ## d= \frac{1}{2} a t^2 ##
    ## t^2 ## is 2 (it looks like he took t as square root of 2).
    So ## d=\frac{1}{2} (\frac{1}{2} \frac{m}{s^2}) \cdot 2s^2 =1/2 m ##
    What I am missing?

    I think that it needs 2s to travel 1 m, at that acceleration. But I may have a brain failure tonight. :smile:
     
    Last edited: Jan 7, 2016
  10. Jan 7, 2016 #9

    HallsofIvy

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    Again, you are assuming that this has initial velocity 0- from rest. That was not given.
     
  11. Jan 7, 2016 #10
    if initial speed was 0.35m/s... it checks out ..
     
    Last edited: Jan 7, 2016
  12. Jan 7, 2016 #11
    Aha, this is how you see it. Thank you for clarification.
    It would be interesting to know where that value for time come from.
     
  13. Jan 7, 2016 #12

    russ_watters

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    No - I don't know what I was thinking (I don't remember - either I didn't bother checking the math or did it wrong myself). You're right that they don't match.
     
  14. Jan 8, 2016 #13
    1 kg with a 1 N motor would arrive at the 2 meter mark at 1.414 seconds.
     
  15. Jan 8, 2016 #14
    Is that correct, though?
    I thought that a 1N motor would exert more Joules per amount of time regardless of distance,
    as long as we are comparing different masses.
    It should take more time (& more Joules) for a 1 N motor to move 10 kg a distance of 1 meter, & less time (& less Joules) to move 1 kg a distance of 1 meter, right?
     
    Last edited: Jan 8, 2016
  16. Jan 8, 2016 #15
    It may, if you choose the right initial speed. But not if it start from rest.

    The "Joules per second" is power.
    P=F*v where F is force and v is magnitude of velocity.
    So you either have a motor with constant force and then the power depends on the velocity. So it won't be constant over time if you have accelerated motion.

    Or you have a constant power motor and then the force won't be constant unless.

    And of course, you can have both changing in time.


    The work done is force times distance (W =F*d) so is the same in both cases.
    It would be a good idea to use the actual equations instead of trying to put them in words or trying to guess different relationships where a simple relation will show you imediately that your are wrong (or right).
     
    Last edited: Jan 8, 2016
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