Question: A particle starts from rest at a distance b(>a) from a fixed point O under the action of a force through a fixed point, the law of which at a distance x from O is $F= \mu (1-a/x)$ towards O when $x>a$ but $F= \mu(a^2/x^2 - a/x)$ away from O when $x<a$. Show that the velocity of the particle will again vanish when $x= a^2/b$
I've taken the differential equations $v dv/dx = -\mu(1-a/x)$, limit $v:0 \rightarrow V$, $x:b \rightarrow x$ and $v dv/dx=\mu(a^2/x^2 - a/x)$, limit $v:V \rightarrow 0$, $x:a \rightarrow x$ After that I'm stuck.
It would be preferable to compare the work done on the particle along each path to $ \ x = a \ . $ For $ \ x > a \ , $ the force acts "to the left", so the change in kinetic energy of the particle is $$ \ -\int_b^a \mu·(1 - \frac{a}{x}) \ dx \ \ = \ \ \mu \ · \ (a \ln x \ - \ x)|_b^a \ \ . $$ When $ \ x < a \ , $ the force acts "to the right", so the change in KE along this path is $$ \ \int_X^a \mu·(\frac{a^2}{x^2} - \frac{a}{x}) \ dx \ \ = \ \ \mu · \left( -\frac{a^2}{x} - a \ln x \right)|_X^a \ \ , $$ where $ \ X < a \ $ is the starting point. Since the particle is (momentarily) at rest at each end, these work integrals will be equal.
While you could solve the equation for $ \ X \ , $ you are just asked to "verify" that setting $ \ X \ = \ \frac{a^2}{b} \ $ makes the two amounts of work equal.