Classical Dynamics: Given v(x), find F(x), x(t), and F(t).

[BlueFalcon]

Homework Statement

The speed of a particle of mass m varies with the distance x as v(x) = (alpha)*x^-n.
Assume v(x=0) = 0 at t = 0.
(a) Find the force F(x) responsible.
(b) Determine x(t) and
(c) F(t)

Homework Equations

Likely:
F = ma

The Attempt at a Solution

I obtain
a(x) = -n(alpha)x^-(n+1)
So
F(x) = ma(x) = -mn(alpha)x^-(n+1)

The back of book claims:
F(x) = -mna*x^-(2n+1)

They use 'a' for the answer, I think they mean alpha, unless a IS alpha...

 

[gabbagabbahey]

Hi BlueFalcon, welcome to PF!

I obtain
a(x) = -n(alpha)x^-(n+1)

Careful, acceleration is the change in velocity with respect to time, not position; you need to use the chain rule:

\frac{d}{dt}v(x)=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}

P.S. In the future, problems like this should probably be posted in the introductory physics forum instead.

 

[BlueFalcon]

Hi BlueFalcon, welcome to PF!



Careful, acceleration is the change in velocity with respect to time, not position; you need to use the chain rule:

\frac{d}{dt}v(x)=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}

P.S. In the future, problems like this should probably be posted in the introductory physics forum instead.

BAH!

I swear I tried that method and got a bunch of warrgarrbllll.

Thanks.

I can't believe I messed it up that bad.

 

[BlueFalcon]

Although, I can't seem to find x(t). Running into the same wargarbl.

 

[gabbagabbahey]

Hint: You have a separable ODE for x(t):

\frac{dx}{dt}=v(x)\implies \int \frac{dx}{v(x)}=\int dt

(Don't forget the constant(s) of integration!)