Velocity as a function of distance [v(x)]

[BitterX]

Homework Statement

a body with mass M moves across a plane with friction

friction constant:
$\mu = \lambda x^2$

the body starts at x=0
with velocity v₀

find at what x
the body stops
and what was the velocity half way there.

Homework Equations

$v^2=v_0^2+2a\Delta x$

The Attempt at a Solution

obviously,
$F(x)=mg\mu = mg\lambda x^2$
so
$a(x)=g \lambda x^2$

so in the equation $v^2=v_0^2+2a\Delta x$
I get
$v^2=v_0^2+2g\lambda x^3$the Question is, can I use this equation? the acceleration is not constant and this equation
depend on the fact that $x=v_0t+ \frac{a}{2}t^2$
and $v=v_0+at$
(and it's not true for non-constant acceleration)

if I cant, how can I integrate the acceleration?
or how do I get v(x)?

Thanks.

EDIT:
I used $a= v\frac{dv}{dx}$
therefore
$vdv=adx$

$\int_{v_0}^{v(x)}{vdv} = g\lambda \int_{0}^{x}{x^2}$

$\frac{1}{2} ( v(x)^2- v_0^2) =\frac{1}{3} g\lambda x^3$

$v(x)^2=v_0^2+\frac{2}{3}g\lambda x^3$

does that seem right?

 

[gulfcoastfella]

I don't see anything wrong with your edited solution.

 

[ehild]

Homework Statement

a body with mass M moves across a plane with friction

friction constant:
$\mu = \lambda x^2$

the body starts at x=0
with velocity v₀

obviously,
$F(x)=mg\mu = mg\lambda x^2$
so
$a(x)=g \lambda x^2$

I used $a= v\frac{dv}{dx}$
therefore
$vdv=adx$

Does the friction increase speed?
Remember that velocity, acceleration and force are all vectors. You need to use proper signs with them.

ehild

 

[BitterX]

ah, of course... it's with a minus :)
on paper I actually did it with a minus. Thanks for pointing it out though!

 

[asteriatic]

Yes, your approach is correct. Integrating the acceleration function will give you the velocity function as a function of distance. Your final equation for v(x)^2 is also correct. To solve for the stopping point, you can set v(x) = 0 and solve for x. To find the velocity at the halfway point, you can plug in x = (1/2)x_max (where x_max is the stopping point) into your equation for v(x)^2.