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Mass subject to a time varying force, find acceleration/speed.

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m=2.2kg is subject to a time varying force of the form

    F(t) = Foe-bt

    Given that Fo = 75N, b = .11*(1/s), particle is at rest at t=0s

    find:

    a) Acceleration of the particle at t=1s
    b) Speed of the particle at t=1s

    Remember that d/dt eat = aeat
    2. Relevant equations
    F=ma

    Kinematics:
    y = Vot + (1/2)at2
    Vf = Vo + at
    Vf2 = Vo2 + 2ay
    y = t * [ (Vo + Vf) / 2 ]

    3. The attempt at a solution
    I'm extremely lost on this.

    I know the first derivative of position is velocity (speed) and the 2nd derivative is acceleration.

    But this just seems to have so much going all at once I'm not entirely sure what to do.

    Does it have anything to do with dv/dt = a = F/m ? My first attempt I tried taking the first and second derivative and solving for t=1s from the equation. Then forgot to do anything about the 2.2kg so I know I got it wrong.

    My second attempt looks like this:
    a) find acceleration at t=1s
    a=dv/dt=F/m
    so F(t)/m = a = (Foe-bt) / m
    Substitute b for .11 1/s and t = 1s and Fo = 75N
    (75N * e-.11*(1/s)(1s))/2.2kg
    The seconds in the superscript cancel out leaving...
    (75N * e-.11)/2.2kg
    N is kg * m / s^2 so the masses cancel out after the division leaving acceleration.
    75/2.2 ~=~ 34.1
    a = 34.1 * e-.11 m/s2 (which is around 30.5 m/s^2)

    b) find speed at t=1s
    The integral of F(t) evalutated at t=1s should be the speed.
    1/m ∫Foe-bt dt
    =
    (1/m)(-1/b * Fo e -bt ) +C
    Substituting and reducing fractions once again leaves me with...
    1/2.2kg * -9.09s * 75 kg*m/s^2 * e-.11 + C
    Mass cancels out, one unit of time cancels out in kg m /s^2. All this leaves us with m/s which we want.

    However this leaves me with a negative number. -277.61 m/s + C. So unless C was to make it positive, this is incorrect.

    Am I on the right track at all?
     
    Last edited: Jan 29, 2014
  2. jcsd
  3. Jan 29, 2014 #2

    CWatters

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    Science Advisor
    Homework Helper

    I haven't checked your working but...

    The integral of acceleration is velocity not speed. I see nothing wrong with the velocity being -ve. In which case the speed is just the magnitude of the velocity.

    You know the velocity at t=0. Does that help you calculate C?
     
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