# Classical mechanics problem

## 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.

## The Attempt at a Solution

give me a clue on how to do this

ideasrule
Homework Helper
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
Homework Helper
Gold Member
Use

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

and find v(x).

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
Homework Helper
Gold Member
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?

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.

Last edited:
kuruman
Homework Helper
Gold Member
How do you get two times from

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

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

Last edited:
Sorry, my bad, the expresion on the right is squared, so theres both negative and positive roots, but negative times are unphysical , il consider the positive root.

kuruman
Homework Helper
Gold Member
Sorry, my bad, the expresion on the right is squared, so theres both negative and positive roots, but negative times are unphysical , il consider the positive root.

How do you get a positive root out of

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

??

Check the math that got you to this point.

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
Homework Helper
Gold Member
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

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

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.