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Find time given a force with respect to displacement

  1. Jun 24, 2017 #1
    1. The problem statement, all variables and given/known data
    With a function that gives Force with respect to Displacement (F(x) = 2000-100x), is there a way to find a general formula for the time it takes from displacement 0 to 20? The force is acting on an object, pushing it horizontally. Assume no friction and no air resistance.

    2. Relevant equations
    Force with respect to Displacement (x):
    F(x) = 2000-100x
    At 0 displacement, there's 2000 Newton of force acting on the object, and the force linearly decreases until there's no force at 20 meters away and beyond.

    3. The attempt at a solution

    Work = ∫ from 0 to b of (F * dx) = ∫ from 0 to b of (2000-100x) dx = -50(b- 40)b
    Work = Final Kinetic Energy - Initial Kinetic Energy. Initially, the object is stationary with no initial velocity, so I'm assuming that the initial Kinetic Energy is 0.

    Work(b) = Kinetic Energy(b) = -50(b - 40)b = 0.5mv^2.

    Solving for v, I get: v(b) = sqrt((2/m) * -50(b - 40)b)

    From here, I'm tempted to plug velocity into this formula (Displacement = Velocity * Time) and solve for time, but I feel like this won't work.
     
  2. jcsd
  3. Jun 24, 2017 #2
    I think you only need Newtons II law to solve this. You get 2. order differencial equation. General formula has mass as variable.
     
  4. Jun 25, 2017 #3

    haruspex

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    That is only true for constant velocity. The general expression is the integral of velocity wrt time. That will give you a differential equation to solve.
    The first integration stage from that equation produces the energy relation Kiwigami obtained, so there is no benefit in going that route.
     
  5. Jun 25, 2017 #4
    I meant integrating over time.
     
  6. Jun 25, 2017 #5

    haruspex

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    How are you going to do that, given acceleration as a function of distance?
     
  7. Jun 25, 2017 #6
    just write Newton's II law(##\frac{\partial^2 x}{\partial t^2}=F/m##) and replace F to function, that he gave. Doing that will result getting second-order linear ordinary differential equation. Also is known that displacement and speed at time 0 are 0.
     
  8. Jun 25, 2017 #7

    haruspex

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    You did not answer my question. How are you going to integrate F, a function of x, with respect to time?
     
  9. Jun 25, 2017 #8
    By writing Newton's II law(##\frac{\partial^2 x}{\partial t^2}=F/m##) and replacing force F to function, that he gave and taking undefined time-integral of both sides. Taking this integral is not necessary.
    ##\frac{\partial^2 x}{\partial t^2}=\frac{2000-100x}{m}## ⇔ ##x+k_1+k_2 \cdot t=\iint(dt^2 \cdot \frac{2000-100x}{m})## .
    by ##x(0)=0## and ##\frac{\partial x}{\partial t}(0)=0## we know that ##k_1=0## and ##k_2=0##.
     
    Last edited: Jun 25, 2017
  10. Jun 25, 2017 #9

    haruspex

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    I ask a third time, how are you going to perform that integral of x wrt time? Writing down the integral doesn't answer it, you actually need to integrate it.
     
  11. Jun 25, 2017 #10
    It (##\frac{\partial^2 x}{\partial t^2}=\frac{2000-100x}{m}##) is called differential equation. Before solving it I cant find that ##\iint(dt^2 \cdot \frac{2000-100x}{m})+k_3+k_4 \cdot t= c_2 \cdot sin(\frac{10 \cdot t}{\sqrt{m}}) + c_1 \cdot cos(\frac{10\cdot t}{\sqrt{m}}) +20##
    , but after solving it I can.
     
  12. Jun 25, 2017 #11

    haruspex

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    But having solved the D.E. by some other method, you do not need to do this integral, so writing it out as a double integral of x wrt t was a complete waste of time. Including mine.
     
  13. Jun 25, 2017 #12

    TSny

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    What if you repeat the above calculation, but you integrate from 0 to some arbitrary value of x between 0 and b? Would you then be able to obtain an expression for the velocity as a function of x?
     
  14. Jun 25, 2017 #13

    Ray Vickson

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    Things simplify if you recognize that you can write ##F = -100(x-20)##, so in terms of the new displacement variable ##y = x - 20## the force law is just ##F = - 100y##. This is Hooke's law (for stretching forces in a spring), so a Google search on "Hookes law" might be very helpful.
     
    Last edited: Jun 25, 2017
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