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Given: F(x)=Fo+ct, Fo is independent of v

Find x dot, and x.

- Thread starter infraray
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- #1

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Given: F(x)=Fo+ct, Fo is independent of v

Find x dot, and x.

- #2

HallsofIvy

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Find x dot, and x."

If the problems are phrased like that, it's no wonder you have problems! There is no "x" or "x dot" in the problem. Also you have "F(x)" but Fo+ ct doesn't depend on x!

I

One very important "problem solving skill" is learning basic definitions and formulas. If you are taking a mechanics class then surely someone has mentioned F= ma! If your object has mass m, then ma= F= F

Now, speed (x dot) is the derivative of position (x) so to find x, you have to find the anti-derivative of that.

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- #4

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xdot=bx^-3

b is positive constant, find force acting on particle as function of x.

I am not sure where to even start. I know F = ma = m(xdot)*(d(xdot)/dx). I think I am getting tripped up on converting the derivative m(xdot)*(d(xdot)/dx) into something I can use on left side of my initial equation. Is this the correct first step?

- #5

TD

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[tex]a = \frac{{d^2 x}}{{dt^2 }} = \ddot x = \frac{{d\dot x}}{{dt}}[/tex]

We know that [tex]\dot x = bx^{ - 3} [/tex]

So

[tex]a = \frac{{d\left( {bx^{ - 3} } \right)}}{{dt}}[/tex]

And

[tex]F = ma = m\frac{{d\left( {bx^{ - 3} } \right)}}{{dt}}[/tex]

- #6

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But shouldn't this be with respect to x instead of t as you have shown as I am looking for F(x).

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TD

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- #8

HallsofIvy

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If [itex]F(x)= m\ddot x[/itex], you will need to use the chain rule: [itex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}[/itex]. But, of course, [itex]v= \frac{dx}{dt}[/itex] so [itex]\frac{dv}{dt}= v\frac{dv}{dx}[/itex]. With v= bx

F(x)= m(bx

- #9

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Just wanted to say thanks for the help. Your pointers helped.

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