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Constant power, resistance, work, energy

  1. Apr 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Mass car 1200 kg .

    power car constant 20kw .

    Resistance car motion 500N.

    Car passes point X and Y with speeds 10 m/s ,25 m/s.

    Car takes 30.5 s to travel from X to Y

    1. Find acc of car at x
    2.considering work/energy find distance XY


    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Apr 25, 2015 #2
    T
    tried a few but looks nowhere near promising and I don't have an answer for the question ,past year examinations
     
  4. Apr 25, 2015 #3
    It is difficult to provide you any guidance if you do not show us your work. Please post your attempt at a solution, no matter how incorrect you may think it is.
     
  5. Apr 25, 2015 #4
    You might start by finding the net driving force at point A ( car driving force = power (watts) / velocity (m/s) )
    The acceleration at that point = net force / mass

    Hint: A velocity v acceleration graph covering the speed range will be a curve
     
  6. Apr 25, 2015 #5
    1. Distance used from question 2.
    work = force x distance
    Force =610KJ/470 m
    =1297.9
    acceleration = force/mass
    =1297.9/1200
    =1.0815

    2. Power x time = work
    Work= 20kw x 30.5
    =610kj

    Resistance = 500N x distancexy

    Net =kinetic energy
    610kj - (500N x distancexy )= (0.5 )
    (1200)
    (25^2)

    Therefore distance xy = 470 m
     
  7. Apr 25, 2015 #6
    This looks like a calculus solution
     
  8. Apr 25, 2015 #7
    By short step excel based start and step program i get around 293 m distance
     
  9. Apr 25, 2015 #8
    Hmm I see I will try to work it out , thanks for the help
     
  10. Apr 25, 2015 #9
    The program consides short velocity changes, adding the derived distance changes as it goes.
    Consider the velocity change between 10 m/s and 11 m/s, calculate the (net) driving force at each velocity, find the average (net) force, calc the acceleration from : a = f net ave / mass, then use newons laws to calculate the distance travelled between those two velocities.
    Add the steps as you go for a running distance.
    Hint: i used steps of 0.01 metres / second, and 1500 ish lines of program
     
  11. Apr 25, 2015 #10
    Sorry am not really sure about calculus part
    But maybe this is the solution

    Driving force =1200*acc - 500N
    P=fv
    20kw=1200*acc - 500N
    Acc=2.083

    2.
    Driving force =1200(2.083)- 500
    =1999.6

    Fs=0.5mv^2 - 0.5 mu^2
    S={ [(0.5)(1200)(25^2)-[(0.5)(1200)(10^2)]} / driving force

    Therefore distance =157.6m
     
  12. Apr 25, 2015 #11

    NascentOxygen

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    Staff: Mentor

    You are almost right.

    net work done = change in KE
    and the car already has some KE at the beginning of that 30.5s time interval.
     
  13. Apr 25, 2015 #12
    I don't think that this is correct, being as the force that propels the car forward is not constant.

    A power output is not a force.
     
  14. Apr 25, 2015 #13
    Ao

    So the 2nd answer i gave should be the correct one ?

    Thanks
     
  15. Apr 25, 2015 #14

    NascentOxygen

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    Staff: Mentor

    No, because you introduced other mistakes there. :frown:

    Fix the one I commented on.
     
  16. Apr 25, 2015 #15
    Yea I thought the 2nd answer was correct because it fixes the change in kinetic energy as you had pointed out
     
  17. Apr 25, 2015 #16

    haruspex

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    you wrote, correctly, P=F v, but that's not what you did in the next step. What happened to v?
     
  18. Apr 25, 2015 #17
    Owh sorry I made a typo I did include the v in the calculation
    V = 10 at x
    20k=(1200*accleration - 500 ) (10 ) Acc= 2.083
     
  19. Apr 25, 2015 #18

    haruspex

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    Subtracting the 500N resistance made the acceleration go up. Does that seem reasonable?
     
  20. Apr 25, 2015 #19
    Thanks got it guys , conceptual problem
     
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