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Homework Help: Work Energy Concept

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    http://xs511.xs.to/xs511/07046/workenergy.JPG [Broken]


    2. Relevant equations



    3. The attempt at a solution

    Ok im all confused on this one.

    ok so leave part (I) and (II)

    onto part (III)

    word done by the pulling force acting on the block = the loss of energy + work done by resisting force

    by loss of energy i mean gain in PE - loss in KE

    =(7500+225)-(1225)+(7.5 x 200) = 8000J

    now im confused as to whether or not the value of theta is mgsintheta or fscostheta please help
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 27, 2007 #2

    radou

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    Don't overcomplicate things. The work done by all the forces acting of the block from A to B equals the change in kinetic energy from A to B. Regarding the trig, draw a sketch.
     
  4. Jan 27, 2007 #3
    ok then the loss in KE=1000J
    but thats not the right answer, theres also resistance forces how do i take that into account and what do you mean by internal and external forces.

    http://xs511.xs.to/xs511/07046/trainagent.JPG [Broken]
     
    Last edited by a moderator: May 2, 2017
  5. Jan 27, 2007 #4

    radou

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    work done by pulling force + work done by resistance force = change in kinetic energy. => (iii) work done by pulling force = change in kinetic energy - work done by resistance force. This should do.
     
  6. Jan 27, 2007 #5
    work done by pulling force = 1000 -(7.5x200)=-500 J

    i though it was

    work done by pulling force= gain in ke + gain in ke + work done by resisting force

    but if its loss in ke then we minus it
     
  7. Jan 27, 2007 #6
    just bump this up.
     
  8. Jan 27, 2007 #7

    arildno

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    Let's take one question at a time:

    How would you calculate the answer to i)?
     
  9. Jan 28, 2007 #8
    green pandas

    loss in KE=1/2m(v^2-u^2)=
    1/2(50)(49-9)=1000J

    Gain in PE=mgh=(50)(10)(15)=7500J

    so has the car goes up the straight hill it gains 7500J in PE but looses 1000J KE.
    Because thers a resistance force of 7.5 N, the work done by it=fs=7.5(200)=1500J

    So work done py pulling force=(7500-1000)+(1500)=8000J
     
  10. Jan 28, 2007 #9
    thats wrong

    work done by pulling force - work done by resistance=change in kinetic energy
     
  11. Jan 28, 2007 #10

    radou

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    If I wrote '+', that doesn't mean that the work done by resistance doesn't have a '-' to reveal behind its name. :wink:
     
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