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

  • Thread starter BitterX
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  • #1
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Homework Statement


a body with mass M moves across a plane with friction

friction constant:
[itex]\mu = \lambda x^2[/itex]

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



[itex]v^2=v_0^2+2a\Delta x[/itex]

The Attempt at a Solution



obviously,
[itex]F(x)=mg\mu = mg\lambda x^2[/itex]
so
[itex]a(x)=g \lambda x^2[/itex]

so in the equation [itex]v^2=v_0^2+2a\Delta x[/itex]
I get
[itex]v^2=v_0^2+2g\lambda x^3[/itex]


the Question is, can I use this equation? the acceleration is not constant and this equation
depend on the fact that [itex]x=v_0t+ \frac{a}{2}t^2[/itex]
and [itex]v=v_0+at[/itex]
(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 [itex] a= v\frac{dv}{dx}[/itex]
therefore
[itex]vdv=adx[/itex]

[itex]\int_{v_0}^{v(x)}{vdv} = g\lambda \int_{0}^{x}{x^2}[/itex]

[itex]\frac{1}{2} ( v(x)^2- v_0^2) =\frac{1}{3} g\lambda x^3[/itex]

[itex]v(x)^2=v_0^2+\frac{2}{3}g\lambda x^3[/itex]

does that seem right?
 
Last edited:

Answers and Replies

  • #2
gulfcoastfella
Gold Member
99
1
I don't see anything wrong with your edited solution.
 
  • #3
ehild
Homework Helper
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1,912

Homework Statement


a body with mass M moves across a plane with friction

friction constant:
[itex]\mu = \lambda x^2[/itex]

the body starts at x=0
with velocity v0

obviously,
[itex]F(x)=mg\mu = mg\lambda x^2[/itex]
so
[itex]a(x)=g \lambda x^2[/itex]

I used [itex] a= v\frac{dv}{dx}[/itex]
therefore
[itex]vdv=adx[/itex]


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

ehild
 
  • #4
36
0
ah, of course... it's with a minus :)
on paper I actually did it with a minus. Thanks for pointing it out though!
 

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