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Newtonian mechanics on a particle

  1. Jan 27, 2004 #1
    Say that a particle of mass m slides down an inclined plane under the influence of gravity. If he motion is resisted by a force f=kmv^2, how canyou show that the time requied to move a distanc d after starting form rest is t= cosh^-1(e^kd)/(sqrt kg sin theta) where theta is the angle f inclination o the plane.
  2. jcsd
  3. Jan 28, 2004 #2


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    When will people learn that it is impossible to give ideas about how to solve some of these problems without knowing what they have to "work with"! That's one good reason to SHOW WHAT YOU HAVE DONE SO FAR and what you do understand about the problem. I would set this up as a differential equation and then solve that equation but that may not be the way you would need to do it.

    The first thing you would do is draw a diagram! You should then see from the diagram that the component of gravitational force (weight) down the slope is -mg cos(θ). Since there is resisting force kmv2, The total force, taking "+" up the inclined plane is kmv2- mg. Since "force= mass times acceleration": mdv/dt= kmv2- mg, a differential equation.
    That's relatively easy to solve (If you have studied differential equations!) but you don't really have to do that. You asked "how can you show that the time requied to move a distanc d after starting form rest is t= cosh^-1(e^kd)/(sqrt kg sin theta)". Since you are given tis you can "turn it around" by solving for d (that's relatively easy since the obvious inverse of cosh-1() is just cosh(). After you have d as a function of t, differentiate to get v as a function of t. Now plug that into the equation and show that it satisfies it.
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