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Is this a constant or variable force?

  1. Apr 6, 2009 #1
    If a vehicle is stopping its change in Kinetic energy is equal to the force times displacement or distance as it is traveling in a straight line and parallel to displacement.

    If the force is F(net) = kinetic + drag

    F(net) = umg + 1/2*A*rho*v^2

    is this force constant or variable

    If it is constant then the Kinetic energy (T) is

    delta(T) = F(net) delta(s)

    If it is variable then it is the formula for variable force

    What I'm hoping to get is a differential equation of the form

  2. jcsd
  3. Apr 6, 2009 #2


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    The force is clearly a function of velocity and hence time and is therefore, non-constant.

    What exactly are you trying to solve for?
  4. Apr 6, 2009 #3
    I have solved it for distance.

    dT/ds = B+CT where v^2=2T/m

    solving I got

    s = ln(B+CT) - ln(B) = ln(B+CT/B)

    looking back through definitions I have noticed that I have used a formula for constant force when it isn't but I'm not sure
  5. Apr 6, 2009 #4


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    I'm not sure why/how you're invoking energy here. And no, as I said above you cannot treat the force as constant.

    A must more straightforward method would be to simply note the definition of force in terms of acceleration and solve the resulting ODE.
  6. Apr 6, 2009 #5
    I have a vehicle stopping. I have already solved this issue and am writing the result up. I have descibed the stopping distance of the vehicle as a differential equation that describes the rate that the vehicle is losing energy. When the vehicle has stopped its Kinetic energy is zero.

    The solution I have for this is

    Where Work = change in Kinetic energy = Force * displacement

    dT = F(net)ds

    dT/ds = F(net)

    where F(net) = umg + drag equation http://en.wikipedia.org/wiki/Air_resistance

    for the v^2 term in the drag equation I substitute V^2 = 2T/m


    F(net) = B + CT

    the frictional force plus the drag force in terms of kinetic energy

    I have also calculated a numerical solution using an Euler method that iterates down using the differential equation.

    The problem I have had is do I get the same result for variable force as I do with constant force
    dT = F(net)ds
  7. Apr 6, 2009 #6


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    You write it as an ODE.

    [tex]dT/ds = -b - cT[/tex]

    Which has the solution,

    [tex]T = (T_0 + b/c) e^{-cs} - b/c \,\,\,\,\,[/tex] : valid while T>0

    BTW. Have you done ODE's in maths yet?
  8. Apr 6, 2009 #7
    yeah I've done one or two. I needed to solve it for s(max). The solution to the problem is how far the car travels not the kinetic energy of the vehicle.

    dT/ds = b+cT

    we get

    dt/(b+cT) = ds


    (b+cT)=x, dx = cT


    T = 1/c(dx)

    we now get

    1/c (dx/x)=ds


    integral b+cT0 - b(1/c){1/x} =integral 0 - smax{s}

    we get

    smax = 1/c{ln(b)-ln(b+cT0}

    The constants are negative.

    is there a way ogf using equation editor

    I just need to know if I can get

    dT = F(net)ds from the variable force if you can help
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