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 :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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