# Position dependant forces

1. Oct 12, 2004

### Speags

Find the velocity v as a function of the displacement x for a particle of mass m, which starts from rest x=0
F(x)=Fo + Cx where Fo and C are positive constants
So far i've gotten,
ma=Fo + Cx
m (dv/dt)=Fo +Cx
m (dv/dx dx/dt)=Fo +Cx I split dv/dt using the product rule
m v dv=(Fo + Cx) dx v=dx/dt
now i'm haveing problems doing the intergal of both sides i have so far
m(v-vo)=????

can anyone help?
I've intergrated on the LHS from vo to v and i think the RHS should be from xo to x

2. Oct 12, 2004

### Pyrrhus

What the problem?

$$mvdv=(F_{o} + C_{x})dx$$

$$\int_{v_{o}}^{v} mvdv= \int_{x_{o}}^{x} (F_{o} + Cx)dx$$

$$\int_{v_{o}}^{v} mvdv= \int_{x_{o}}^{x} F_{o}dx + \int_{x_{o}}^{x} Cxdx$$

Mass is a constant
Fo and C are positive constants

Integrate.

$$m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C \int_{x_{o}}^{x} xdx$$

You should have:

$$m \frac{1}{2}v^2]_{v_{o}}^{v}= F_{o}x]_{x_{o}}^{x}+ \frac{1}{2}Cx^2]_{x_{o}}^{x}$$

$$\frac{1}{2}m(v^2 - v_{o}^2)= F_{o}(x-x_{o})+ \frac{1}{2}C(x^2-x_{o}^2)$$

Our Particle starts from rest (Vo = 0) at x = 0 so

We got

$$\frac{1}{2}m(v^2)= F_{o}x+ \frac{1}{2}Cx^2$$

$$v(x) = \sqrt{\frac{2F_{o}x+ Cx^2}{m}}$$

There you go.

Last edited: Oct 12, 2004
3. Oct 12, 2004

### Speags

$$m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C_{x} \int_{x_{o}}^{x} dx$$
should be

$$m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C \int_{x_{o}}^{x} xdx$$
sorry i didn't explain it well enough C and X are seperate

so does that mean the solution would be:
$$1/2mv^2=F_{o}x + 1/2Cx^2$$
$$v= \sqrt {(2F_{o}x)/m + (Cx^2)/m}$$

4. Oct 12, 2004

### Pyrrhus

Oh ok no problem, but at least you got it

5. Oct 12, 2004

### Speags

$$v_{o} = 0 , x_{o} = 0$$