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COnversative Forces and Change in Kinetic Energy

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A roller coaster (365. kg) moves from A (5.00 m above the ground) to B (28.0 m above the ground). Two nonconservative forces are present: friction does -2.00 multiplied by 104 J of work on the car, and a chain mechanism does +3.00 multiplied by 104 J of work to help the car up a long climb. What is the change in the car's kinetic energy, ΔKE = KEf - KE0, from A to B?

    2. Relevant equations
    change in kinetic energy =work or final KE minus initial KE.
    Work=force x distance
    change in KE= .5(m)(final velocity)^2

    3. The attempt at a solution
    I used the constant acceleraton euqation to solve for final velocoty. And the formula .5(m)(final velocity)^2 to calculate KE.
    I got 10000. This answer was wrong.
  2. jcsd
  3. Jan 27, 2010 #2
    You are missing another type of work here: the work done against the coaster by Earth's gravity. Plus, there's never any need to use acceleration or velocity in this question; it is purely Work/Energy.
  4. Jan 27, 2010 #3
    so how would i incorporate earth's gravity 9.8?
  5. Jan 27, 2010 #4
    9.8 is acceleration, not work. As you wrote, work is force multipled by distance.
  6. Jan 27, 2010 #5
    Ep=Ek Ep=mgh (mass*gravity*height) Ek=1/2mV2 V is velocity at that point you do not need to find final velocity you must manipulate the equations mgh=1/2*mv2 = [STRIKE]m[/STRIKE]gh=1/2* [STRIKE]m[/STRIKE] v2 m cancels out you just need to manipulate the equation and you can find anything
  7. Jan 27, 2010 #6
    okay but what do i do about the two nonconservative forces given that -2x10^3 is work done by earth
  8. Jan 27, 2010 #7
    Ff as i know it is equal to Fn * [tex]\mu[/tex] [tex]\mu[/tex]= to the kinetic force of friction. NOTE: there are 2 types of friction static and kinetic static is just simply what the friction force is when the object is at rest kinetic friction is what most people find because it is what is most commonly used and referred too (or so i have been told) i dont think this answer your question but i think it should help cause i was think along the lines of wtf? what does energy have to do with this but looking at it again i realize that one could manipulate the equation so that it may be a vector with velocity if i might ask what level of physics is this?
    Last edited: Jan 27, 2010
  9. Jan 27, 2010 #8
    You've already got all you need, forget about the force of friction. You've got +30,000 J of work done by the chain, -20,000 J of work done by friction, and you've got to figure out the work done by gravity (-2,000 J is not correct for that). Then just figure out how much all this work adds up to, and you're done!
  10. Jan 27, 2010 #9
    that is the problem. i don't understand work by gravity!
  11. Jan 27, 2010 #10
    this is high level college physics
  12. Jan 27, 2010 #11
    what is the force of gravity on the coaster, and does it point up or down? how far does the rollercoaster move while under the influence of this force, and does it move up or down? Work is force times distance, and is either positive or negative depending on the directions of force and distance, and there you go.
  13. Jan 27, 2010 #12
    lol i'm about to write my grade 11 level final exam for physics in about 14 hrs and this stuff almost makes total sense
  14. Jan 27, 2010 #13
    it points down.
    distance is 23
  15. Jan 27, 2010 #14
    so after i find the fwork what do i do?
  16. Jan 27, 2010 #15
    i still don't get this
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