How Can Viscous Drag Affect the Velocity of a Sliding Block?

[Mechdude]

Homework Statement

heres a problem i working on
a block slides on a horizontal surface which has been lubricated with a heavy oil such that the block experiences a viscous drag
resistance in the opposite direction to the velocity and that varies with speed according to:
$$F\left( v\right) =-c\,i\,{v}^{n}$$
if the initial speed is $v_0$ at $t=0$
a) find $v$ as a function of time $t$
which iv found ;
$${\left( {v_o}^{1-n}+\frac{c\left( n-1\right) \,t}{m}\right) }^{\frac{1}{1-n}}$$
b) For $n = \frac{1}{2}$ find the displacement x$\vec{i}$ as a function of t: i is the unit vectoer in x direction. .
which iv found
$$x- x_o = \frac{12\,{m}^{2}\,t\,vo+6\,c\,m\,{t}^{2}\,\sqrt{vo}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}$$

c) show that for $n = 1/2$ , the block will not travel further than $$\frac{2mv_o^{\frac{3}{2}} }{3c}$$
now this is the sucker that will not work out. the answers for a and b are as given in the
assignment sheet, so they are ok.

Homework Equations

The Attempt at a Solution

give me a clue on how to do this

 

[ideasrule]

I really don't think b) can be right. For one thing, x-x0 increases without bound as t increases, which implies the block will eventually reach infinite speed.

 

[kuruman]

Use

$$a=v\frac{dv}{dx}$$

and find v(x).

 

[Mechdude]

I really don't think b) can be right. For one thing, x-x0 increases without bound as t increases, which implies the block will eventually reach infinite speed.

correction
the second equation that i found has a sign problem on the second term compared to the solution provided
$$x- x_o = \frac{12\,{m}^{2}\,t\,vo-6\,c\,m\,{t}^{2}\,\sqrt{v_o}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}$$

 

[kuruman]

OK, do it your way. Can you find the velocity as a function of time from your expression? Can you find at what time that velocity goes to zero?

 

[Mechdude]

OK, do it your way. Can you find the velocity as a function of time from your expression? Can you find at what time that velocity goes to zero?

so from the espression for $f(v)= -cv^n$we can get using$f=ma$ one of Newtons laws
$a= dv/dt=\frac{-cv^n }{m}$
and a little separation of variables,
$$v(t)= {\left( {v_o}^{1-n}+\frac{c\left( n-1\right) \,t}{m}\right) }^{\frac{1}{1-n}}$$
the integration was done much like that of obtaining Newtons equations of motion, from 0 to t, and $dv = v -v_o$ but since the force is variable
the acceleration was not treated as a constant.
integrating this wrt time again we get
$$x-x_o= \frac{t\,\left( 12\,{m}^{2}\,v_o-6\,c\,m\,t\,\sqrt{v_o}+{c}^{2}\,{t}^{2}\right) }{12\,{m}^{2}}$$
and we get this expression from v(t) at n=1/2
$$v(t)={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}$$
when the block stops moving v will be =0 hence we can solve for t at v=0
$$t= \pm \frac{2\,m\,\sqrt{v_o}}{c}$$
inserting this into the equation for displacement x below that we found
$$\frac{t\,\left( 12\,{m}^{2}\,v_o-6\,c\,m\,t\,\sqrt{v_o}+{c}^{2}\,{t}^{2}\right) }{12\,{m}^{2}}$$
we get after a little algebra (for the negative t, il try the positive t later)
$$-\frac{14\,m\,{v_o}^{\frac{3}{2}}}{3\,c}$$

regards
mechdude.

 

[kuruman]

How do you get two times from

$$0={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}$$

*** Added on Edit ***

Also, negative times are unphysical. Check your signs along the derivation of the expression above.

 

[Mechdude]

Sorry, my bad, the expresion on the right is squared, so there's both negative and positive roots, but negative times are unphysical , il consider the positive root.

 

[kuruman]

Sorry, my bad, the expresion on the right is squared, so there's both negative and positive roots, but negative times are unphysical , il consider the positive root.

How do you get a positive root out of

$$<br /> 0={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}<br />$$

??

Check the math that got you to this point.

 

[Mechdude]

starting from the expression for velocity;
$$v(t)= {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) }^{2}$$
and expanding it $$v(t)= v_o-\frac{c\,t\,\sqrt{v_o}}{m}+\frac{{c}^{2}\,{t}^{2}}{4\,{m}^{2}}$$
now if we set v to zero (when it finally stops)
$$0 = v_o-\frac{c\,t\,\sqrt{v_o}}{m}+\frac{{c}^{2}\,{t}^{2}}{4\,{m}^{2}}$$
and solve for t: using the quadratic formula:
$$x=-b \pm \frac{\sqrt{{b}^{2}-4\,a\,c}}{2\,a}$$
$$t=\frac{2\,m\,\sqrt{v_o}}{c}$$
if we substitute this time into the expression for displacement:
$$x-x_o= \frac{12\,{m}^{2}\,t\,v_o-6\,c\,m\,{t}^{2}\,\sqrt{v_o}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}$$

we then get $$x-x_o = \frac{2\,m\,{v_o}^{\frac{3}{2}}}{3\,c}$$
which is what was required.

 

[kuruman]

starting from the expression for velocity;
$$v(t)= {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) }^{2}$$

That's better. Note that in the previous postings you had a minus sign in front of both terms between parentheses. Also, it is much quicker to take the square root on both sides

$$0=\pm {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) }$$
which clearly has one solution

$$t=\frac{2\,m\,\sqrt{v_o}}{c}$$

 

[Mechdude]

That's better. Note that in the previous postings you had a minus sign in front of both terms between parentheses. Also, it is much quicker to take the square root on both sides

$$0=\pm {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) }$$
which clearly has one solution

$$t=\frac{2\,m\,\sqrt{v_o}}{c}$$

thanks, i was not exactly cautious while working the problem,
regards.