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^nwe can get usingf=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

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

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

<br /> t=\frac{2\,m\,\sqrt{v_o}}{c} <br />

 

[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

<br /> t=\frac{2\,m\,\sqrt{v_o}}{c} <br />

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