friction question

**sjangster** · February 9th 2010, 02:58 PM

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?

**felper** · February 9th 2010, 03:04 PM

Remember that f=ma and integrating a twice, you obtain the displacement

**sjangster** · February 9th 2010, 03:06 PM

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?

**HallsofIvy** · February 11th 2010, 02:24 AM

Originally Posted by sjangster

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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