Limit Calculation for Block Sliding on Lubricated Surface with Quadratic Drag

[Jukai]

Homework Statement

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.

Homework Equations

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

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?

 

[fzero]

Homework Equations

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

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

The viscous force acts opposite to the direction of motion. It was already defined with the correct sign, but you negated it. If you take your solution and replace c -> -c you might get the correct result.

 

[Jukai]

The viscous force acts opposite to the direction of motion. It was already defined with the correct sign, but you negated it. If you take your solution and replace c -> -c you might get the correct result.

I just tried it, and my limit is still wrong. My second integral is now x = m/c((1/Vo^(1/2)) - (ct/m))

 

[fzero]

You're missing some minus signs and factors of 2. You're also missing the t=0 term in the integration of the velocity. For the first integration I find

[tex]-\frac{2}{\sqrt{v}} + \frac{2}{\sqrt{v_0}} = - \frac{ct}{m}[/tex]

The 2nd integration gives

[tex] x = \frac{2m v_0 }{c\sqrt{v_0}} \left( - \frac{1}{1+\frac{c\sqrt{v_0}}{2m}t} + 1\right).[/tex]

After simplifying, you get something that yields

[tex]x(t\rightarrow\infty) \rightarrow \frac{2m\sqrt{v_0}}{c}.[/tex]

 

[Jukai]

thank you very much, I understand now