(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A block of mass m slides on a horizontal surface that's been lubricated with a heavy oil so that the block suffers a viscous resistance that varies as the 3/2 power of the speed.

If the initial speed of the block is Vo at x=o, show that the block cannot travel further than 2m(Vo^(1/2))/c

c is the drag constant.

2. Relevant equations

The viscous resistance is defined as F(v)=-cv^(3/2)

3. The attempt at a solution

So I defined my axis so that F = ma = -F(v) = cv^(3/2)

It's obvious that I need to calculate the limit of t when t goes to infinity from the position equation. So I integrate my F=ma equation twice and I don't get the answer.

For my first integration, I have dv/(v^(3/2)) = (cm)dt which gets me 1/(V^(1/2)) - 1/(Vo^(1/2)) = cmt

Now I isolate V=dx/dt in order to integrate a second time. I find that V=1/((Vo^(1/2)) + (ct/2m))^2

Let u = the denominator, du = cdt/2m --> dt = 2mdu/c. The integral becomes dx=1/(u)^2du

and I get x = -1/u. After I replace u for it's value, my limit when t-->infinity isn't what the question demands.

Could someone tell me where I might have made a mistake?

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# Homework Help: Quadratic drag as v^3/2

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