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(simplified?) Air drag force question.

  1. Nov 12, 2013 #1
    My issue/knowledge
    Hello all, I think this is my first time posting a question here (I have read the guidelines). This problem is not in my text book or notes, and I'm not sure how to go about it. Online all I can find about air drag equations requires knowing variables which I'm not given here, so I'm having a great deal of difficulty (for 6 hrs+ now). I'm sorry if my attempts look weak too you, I have really tried. Thanks.

    1. The problem statement, all variables and given/known data

    "A 1500kg car accelerates in first gear to from 0 to 55km/h in 2.7 seconds. In fifth gear from 150 km/h to 220 km/h in 23 seconds. Assume the drag force is negligible at low speeds but increases as the square of the velocity. Estimate the drag force of friction at a speed of ~200 km/h"

    2. Relevant equations

    v^2?
    F=m*a
    a=F/m
    a=v/t
    fk=ukmg
    uk= a/g
    3. The attempt at a solution

    a=v/t
    (55km/h-0km/h) / 2.7 seconds= 5.658 m/s^2 (converted to metric)
    (220km/h-150km/h) / 23 seconds= 0.845 m/s^2
    m*a
    1500kg*5.658m/s^2= 8487N
    1500kg*0.845 m/s^2= 1267.5N

    So I get here and have no idea how to proceed. What I feel like maybe I should do is imagine it like a normal friction problem and just find acceleration it takes to 200km/h, divide by gravity, uk=a/g. Then again I'm thinking that square of velocity is really important, so maybe subtract force of car it takes to get to 200km/h from 200km/h^2, then divide by mass to get acceleration, divide that by gravity? I feel like I'm talking nonesense now.
     
  2. jcsd
  3. Nov 12, 2013 #2
    Gravity plays no role here.

    What does play a role is the force on the car due to its engine, and the force of drag. How are they related?
     
  4. Nov 12, 2013 #3
    Well force of drag goes opposite direction of force of car, so in order to accelerate to higher velocities more force car is required to achieve that acceleration? Hm
     
  5. Nov 12, 2013 #4
    What is the relationship among the force due to the engine, the drag, and acceleration?
     
  6. Nov 12, 2013 #5
    Has to be mass right? Maybe velocity is a better answer.
     
  7. Nov 12, 2013 #6
    "Relationship" means "equation" or "equations"; "mass" is a number.
     
  8. Nov 12, 2013 #7
    Oh sorry. Well a=f/m , and both force due to engine and drag are forces so f=a*m
     
  9. Nov 12, 2013 #8
    What about this?

    So if square of velocity is the force of drag, then (200km/h)^2--->(55.55m/s)^2= 3086.4N
    Then since apparently engine force is 1267.5N, 1267.5N-3086.4N= -1818.9N

    I know I'm probably wrong...If I divided by mass that would give acceleration but I don't think that is drag force of friction. *sigh*
     
  10. Nov 12, 2013 #9
    What makes you think that the engine force is 1267.5 N?

    Note the problem said "Assume the drag force is negligible at low speeds".
     
  11. Nov 12, 2013 #10
    Assuming 0-55km/h is low speed and 150 to 220km/h is high speed, so I calculated the acceleration from 150 to 220 multiplied it by 1500kg after conversion and got that force.

    edit: 0-55 in 2.7 seconds is a quick acceleration hinting that there is negligible drag force...The only data left is 150 to 220 in 23 seconds which is slow hinting drag force.
     
    Last edited: Nov 12, 2013
  12. Nov 12, 2013 #11
    The drag force is NOT negligible at high speeds. 1267.5 N is the sum of the engine and drag forces.
     
  13. Nov 12, 2013 #12
    I see that now wow. I was, still am confused by "increases by square of velocity".
     
  14. Nov 12, 2013 #13
    So, what is the force of the engine at the low speed?
     
  15. Nov 12, 2013 #14
    8487n.
     
  16. Nov 12, 2013 #15
    Assuming the engine produces the same power at the high speed, what is the engine force at the high speed?
     
  17. Nov 12, 2013 #16
    Tried employing work energy theorem but I'm not there yet so....trying to figure out how to do it another way, difficult for me heh.
     
  18. Nov 12, 2013 #17
    Might be useful: Power = Force x Velocity.
     
  19. Nov 12, 2013 #18
    So alright here is what I'm doing. 8487N*(15.277m/s)=129662.5P

    If the power is the same at a velocity of 220km/h->61.11m/s
    Then I can do F=P/velocity, but that gives me =2121.788N? Which is higher than 1267.5N the sum of engine and drag. Is that to say engine power is 2121N and drag is the subtraction result which is -854.288?

    Feel like I'm getting something seriously wrong but I never did these equations before.
     
  20. Nov 12, 2013 #19
    It seems to me that this problem is very badly formulated.

    Using the work-energy theorem.

    The initial velocity is zero and the final velocity is 55 km/h, hence the change in kinetic energy is 175 kJ, divided by 2.7 s, that gives 65 kW of power.

    At the high speed, the initial speed is 150 km/h, the final is 220 km/h, the energy diff is 1.5 MJ. The work of the motive force is 65 kW over 23 s, yielding 1.5 MJ, too. This suggests that the work of the force of drag is zero, which means that there is no drag at the high speed, which obviously is nonsensical. Please check that the numbers you have are copied correctly.
     
  21. Nov 12, 2013 #20
    "A 1500kg car accelerates in first gear to from 0 to 55km/hour in 2.7 seconds. In 5th gear the car accelerates from 150km/h to 220km/h in 23 seconds. Assume the drag force from wind is negligible at low speeds but increases as the square of the velocity. Assuming there is negligible wind drag in 1st gear, estimate the drag force of friction at a speed of ~200km/h?"

    Been staring at this thing for so long now, but I recopied just now. Only difference is that I left "Assuming there is negligible wind drag in 1st gear" out because it seemed rhetorical. Otherwise yeah everything is exact.
     
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