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Math Help - friction question

  1. #1
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    friction question

    1. a block with mass m slides on a horizontal surface. the block is intially at x=0 with velocity v0.

    the frictional force that the block experience is proportional to v^(3/2)

    such that
    F(v)=-kv^(3/2)

    for positive constant K. what is the max distance that the block travel?

    2. a particle of mass m is under the influences of a force F(x) that attracts the particle according to

    F(x) = -kx^(-2)

    where x is the distance from a fixed origin at x=0 and k is a positive constant

    if the particle is released from the rest at x=d, how long does it take to reach the origin?
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  2. #2
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    Remember that f=ma and integrating a twice, you obtain the displacement
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  3. #3
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    yeah i have been doing that but the calculation goes messy and i don't achieve the equation that enables me to get the final displacment

    by doing that i get "t" as denominator meaning the max is infinite

    any help?
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  4. #4
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    Quote Originally Posted by sjangster View Post
    1. a block with mass m slides on a horizontal surface. the block is intially at x=0 with velocity v0.

    the frictional force that the block experience is proportional to v^(3/2)

    such that
    F(v)=-kv^(3/2)

    for positive constant K. what is the max distance that the block travel?
    So F= m\frac{dv}{dt}= kv^{3/2} and [mat]v^{-3/2}dv= \frac{k}{m}[/tex]. You can integrate that to find v(t), determining the constant of integration by the condition that v(0)= v_0. You can then set v(t)= 0 to find t_1> 0 when it stops. Integrate again to find x(t), using x(0)= x_0 to determine the constant of integration, and calculate v(t_1)

    2. a particle of mass m is under the influences of a force F(x) that attracts the particle according to

    F(x) = -kx^(-2)

    where x is the distance from a fixed origin at x=0 and k is a positive constant

    if the particle is released from the rest at x=d, how long does it take to reach the origin?
    Now you have m\frac{dv}{dt}= -kx^{-2}. That has three variables, t, x, and v. But you can use a very handy method called "quadrature" to reduce that.

    By the chain rule, \frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}. But \frac{dx}{dt}= v so \frac{dv}{dt}= v\frac{dx}{dt}.

    Now the differential equation is mv\frac{dv}{dx}= -kx^{-2} or v dv= -\frac{k}{m}x^{-2} dx. Integrate to find v as a function of x, using the fact that x(t= 0)= d, v(t= 0)= 0 which is the same as v(x=d)= 0, to determine the constant of integration.

    Then solve the (separable) first order differential equation \frac{dx}{dt}= v(x) with x(0)= d.
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