Newton's Second Law Particle Problem

[skyrolla]

Homework Statement

4. A particle of mass m is experiences a Force that attracts the particle following
F(x) = -k*x^-2
(an inverse square relationship) where x is the distance from an origin at x = 0 and k is a
positive constant. If the particle is released from rest at x = d, how long does it take to reach the origin?

Homework Equations

F = ma = m(dv/dt)

The Attempt at a Solution

Equating Newton's second and the Force equations yields -k*x^-2 = m(dv/dx)(v).

After seperating the variables and integrating once, I got V = sqrt(2k/m)*sqrt((1/x)-(1/d)).

My boundary condition was that at x = d, v =0 for the integration. Setting V = dx/dt and trying to solve for x(t) is unfathomable, so I'm not sure how to continue here, or if I have been taking the wrong approach. I've double checked my math as well, so I don't believe the error is there, but I could always be wrong.

Thanks

 

[rl.bhat]

Let the velocity of the particle be v when it passes the origin.
The kinematic equation becomes
v = vo + at. But a = -k/m*1/x^2 and vo = 0
So
dx/dt = -k/m*1/x^2 or
x^2*dx = -k/m*t*dt
Find integration between the limits x = d to x = 0 and find t.

 

[tiny-tim]

Hi skyrolla!

(have a square-root: √ and an integral: ∫ and try using the X² tag just above the Reply box )

If you need to solve ∫ √x/(√x - a) dx, just make the obvious substitution.

 

[skyrolla]

I thought that kinematic equation was only applicable to a constant force, not a position-dependent varying one?

 

[rl.bhat]

I thought that kinematic equation was only applicable to a constant force, not a position-dependent varying one?

In the kinematic equation we have taken into account the position dependence of the acceleration. And we have taken the integration to find t. So it must work.