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What equations would I use to solve for work and power delivered?

  1. Oct 4, 2011 #1
    A 6 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 6 + 4x - 3x2, where Fx is in newtons and x is in meters.

    (a) Find the work done by the force as the particle moves from x = 0 to x = 2 m.
    in J
    (b) Find the power delivered to the particle when it is at x = 2 m.
    in W
     
  2. jcsd
  3. Oct 4, 2011 #2

    Pengwuino

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    How do you determine the work done on an object given a certain force acting over a certain distance? This should be a very basic equation that you should know the equation for.

    Also, the units seem to be a bit odd, are you sure that's the correct form of the force?
     
  4. Oct 4, 2011 #3
    fx is 6+4x-3x^2 sorry about that. and would i use the equation w= integral of fx from distance initial to distance final?
     
  5. Oct 4, 2011 #4

    Pengwuino

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    Yes. There you go. Your integration bounds are given the initial and final distances.
     
  6. Oct 4, 2011 #5
    ok now do i have to integrate the fx or is that already accomplished and i should just plug in the numbers, because when i integrate that i get -x^3+2x^2+6x and when i integrate from 0 to 2 i get the answer incorrect.
     
  7. Oct 4, 2011 #6

    Pengwuino

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    No, you must integrate the function. Check your work, you integrated correctly.
     
  8. Oct 4, 2011 #7
    and also how can i find power without a time given
     
  9. Oct 4, 2011 #8
    For that after i integrate i get 20 is that incorrect?
     
  10. Oct 4, 2011 #9

    Pengwuino

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    No, that's not correct.

    I didn't even notice the part asking about power. I really think you haven't written down the problem correctly. Are you sure you don't mean [itex]x(t) = 6 + 4t - 3t^2[/itex]?
     
  11. Oct 4, 2011 #10
    that is the exact copy of my problem. i just looked it over and i am not sure where i am going wrong when i integrate all i should have to do is plug 2 in as x correct because 0 will just yield 0 when plugged in as x
     
  12. Oct 4, 2011 #11

    Pengwuino

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    You don't have enough information to determine the power.

    When you integrate, you should get the work done = 12J, not 20J (check your algebra). I highly doubt you have copied the question correctly. There is no way to determine the power with what you are given.
     
  13. Oct 4, 2011 #12
    that is what i thought and i just double checked that. when you integrate is it products of time inital minus time final? is that where i am making my mistake. I have not integrated in a very long time.
     
  14. Oct 4, 2011 #13
    never mind i realize what i did wrong in integration. stupid mistake. I wish i could figure out the power though
     
  15. Oct 4, 2011 #14
    thanks for your help. i appreciate it. i just gave up on the power. if you could not figure it out i knew it must not be possible.
     
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