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Work done by a parachute

  1. Mar 20, 2012 #1
    i'm trying to figure out the work done by a parachute to slow down a car and so far i have come up with:

    W = F[itex]_{d}[/itex] * d
    F[itex]_{d}[/itex] = C[itex]_{d}[/itex] ρ A[itex]_{f}[/itex] [itex]V^{2}[/itex]/2

    and i have equated the loss of kinetic energy to the work done by the parachute, since the velocity is not constant and therefore the work and kinetic energy are not constant i end up with the following:

    ∫C[itex]_{d}[/itex] ρ A[itex]_{f}[/itex] V dV * d = ∫m V dV

    does this mean that the distance to slow down the car from one velocity to another is independent of velocity? surely the distance to go from 1000mph to 100mph is not the same as the distance to go from 200mph to 100mph

    thanks in advance
     
  2. jcsd
  3. Mar 24, 2012 #2

    Redbelly98

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    Welcome to Physics Forums.

    Equating W with the change in kinetic energy should be enough, shouldn't it? Just plug in the mass and initial velocity of the car to get the initial kinetic energy, and that is equal to the work done to stop the car.
    I disagree with the left-hand side of this equation. The right-hand side is the integral of m·a·dx=F·dx, so that's the work done. The left-hand side is (the integral of dF)*d, so that's the change in force ΔF, times the distance traveled d. And since the force is zero once the car has stopped, that integral is equivalent to the initial drag force on the car. So you have basically said
    (Initial force)·d = work done,​
    which is not true, since the force is continually changing as the car slows down.
     
  4. Mar 24, 2012 #3

    rcgldr

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    Fd is a function of velocity. Acceleraion = f/m so you can change this to

    a = dv/dt = - c v2.

    where c is all of those constants combined.

    To solve this for velocity versus time via direct integration (don't forget to add a constant term for initial velocity at t=0 after integrating):

    dv/dt = - c v2.

    dv/(- c v2) = dt

    Then you can integrate that equation to solve for distance versus time, using the same method.
     
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