Solve the differential equation F=F0+kv

OmegaKV
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Homework Statement



Find the velocity of v as a function of displacement x for a particle of mass m which starts from rest at x=0 and subject to the following force:

[tex]F=F_0+kv[/tex]

You could say mv = F0*t + kx, but the answer in the back of the book is an equation that is only in terms of x and v, not t. The answer in the back of the book involves ln.

Homework Equations



Maybe this:

[tex]\ddot {x}= \frac{d \dot{x}}{dx}[/tex]

The Attempt at a Solution



[tex]m\ddot{x}=F_0 +k\dot{x}[/tex]
[tex]m\dot{x} \frac{d\dot{x}}{dx}=F_0 +k\dot{x}[/tex]
[tex]m\dot{x} d\dot{x} = F_0dx + k\dot{x} dx[/tex]
[tex]\frac{1}{2}m\dot{x}^2 = F_0*x + ?[/tex]
 
Last edited:
on Phys.org
The second equation you wrote down in your attempted solution is separable, but in order to separate it the side with dx should not depend on ##\dot x##.
 
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OmegaKV said:

Homework Statement


Find the velocity of v as a function of displacement x for a particle of mass m which starts from rest at x=0 and subject to the following force:
[tex]F=F_0+kv[/tex]
You could say mv = F0*t + kx, but the answer in the back of the book is an equation that is only in terms of x and v, not t. The answer in the back of the book involves ln.

Homework Equations


Maybe this:
[tex]\ddot {x}= \frac{d \dot{x}}{dx}[/tex]
Is this eq'n dimensionally correct?

The Attempt at a Solution


[tex]m\ddot{x}=F_0 +k\dot{x}[/tex]
[tex]m\dot{x} \frac{d\dot{x}}{dx}=F_0 +k\dot{x}[/tex]
[tex]m\dot{x} d\dot{x} = F_0dx + k\dot{x} dx[/tex]
[tex]\frac{1}{2}m\dot{x}^2 = F_0*x + ?[/tex]
 
rude man said:
Is this eq'n dimensionally correct?
Well, he is not actually using that equation. He is using ##\ddot x = \dot x \, d\dot x/dx##.
 
OmegaKV said:

Homework Statement



Find the velocity of v as a function of displacement x for a particle of mass m which starts from rest at x=0 and subject to the following force:
[tex]F=F_0+kv[/tex]

The Attempt at a Solution


[tex]m\ddot{x}=F_0 +k\dot{x}[/tex]
Assuming F0 and k are constants, how about a substitution of variables to reduce the 2nd order linear ODE into a 1st, then taking orodruin's hint to employ separation of variables to solve the new equation?
 
Orodruin said:
Well, he is not actually using that equation. He is using ##\ddot x = \dot x \, d\dot x/dx##.
Right, but he wrote it down & should learn to check for dimensional consistency, a powerful error-detecting tool that doesn't seem to be sufficiently emphasized in our classrooms.
 

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