Trouble finding F(x) for a particle moving on a horizontal plane

AI Thread Summary
A particle of mass m moves on a frictionless horizontal plane with speed v(x) = k/x. The force acting on the particle is derived from the acceleration, leading to the expression F(x) = -mvk/x^2. Upon substituting v with k/x, the final force function is simplified to F(x) = -mk^2/x^3. Participants in the discussion confirm the correctness of the calculations and share tips on handling calculus concepts. The conversation highlights the importance of expressing force as a function of position.
astenroo
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Homework Statement



A particle of mass m moves along a frictionless, horizontal plane with a speed given by v(x)=k/x, where x is its distance from the origin and k is a positive constant. Find the force F(x) to which the particle is subject.

Homework Equations



F(x)=m\ddot{x}
\ddot{x}=\frac{d\dot{x}}{dt}=\frac{dx}{dt} \frac{d\dot{x}}{dx}=v\frac{dv}{dx}

The Attempt at a Solution


So i figured, if i differentiate v(x)=k/x I end up with v'(x)=-k/x^2, so F(x)= m\ddot{x} and if I substitute my expression for \ddot{x} I end up with F(x)=-mvk/x^2. Why do I get the feeling I'm all wrong about this one?
 
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hi astenroo! :smile:
astenroo said:
… I end up with F(x)=-mvk/x^2. Why do I get the feeling I'm all wrong about this one?

dunno :confused:

looks ok … now get rid of the v, to make F a function of x :smile:
 
Remember that v = k/x. Plug that back into your expression for F(x).
 
Heh, I actually suck at calculus, so when I end up with a fairly straight forward solution, I get the feeling I've done something wrong :) Calculus cannot be simple :D

So, my first attempt was F(x)=-mvk/x^2, and if v(x)=k/x then F(x)=-mk^2/x^3.

Thanks for the hints.
 
btw, note that vdv/dx can also be written d/dx (1/2 v2) :wink:
 
tiny-tim said:
btw, note that vdv/dx can also be written d/dx (1/2 v2) :wink:

Heh, so true. Thank you :)
 
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