Physics problems

Just 2 problems which I think I did right but the anwswer says otherwise. Would any kind anonymous tell me what I did wrong?

1) Potential energy of a system is given as U = 5.6/r^2 - 89/r (in J) , where r > 0 . Calculate the value of r for the equilibrium position.

Student response: 0.07
Correct answer:    0.13

2) A chain of lenth L = 1.4 m and mass m=0.7kg is placed on a smooth table and kept at rest while a piece of length L 1 = 0.7 m is hanging free. After the chain is released it slides from the table. Find the speed of the chain at the moment when the chain leaves the surface of the table.

Student response: 3.2
Correct answer:    3.80

u=1/(r-(5.6/89))r

Constants...
length of chain = 1.4
mass of chain = 0.7
gravity = 9.8

Variables...
length ofchain hanging = L ≤ 1.4
initial L = 0.7
force acting on chain = F = MG = mass of chain hanging * 9.8 = mass of chain * proportion of chain hanging * 9.8 = (9.8*0.7L)/1.4
acceleration = A = F/M = F/0.7 = 9.8L/1.4 = 7L

you need to find the rate of change of acceleration then differentiate twice

OP here.

For the first one I thought that when the potential energy is in an equilibrium position that means that potential energy is equal to zero. Thus I set U = 0 and then solved for r, which I got to be 0.07 ... am I breaking some fundamental laws of physics here?

For the second one I thought that the whole chain is accelerated by the part thats hanging down thus:
m*(dv/dt) = (m/L)*x*g* ; where x = the part that is hanging.

I rewrote dv/dt as (dv/dx)(dx/dt) = v(dv/dx) thus my first equation is now:
vdv = (g/L)xdx
I then intergrated that equation and got:
v^2 = (g/L)(L^2-x^2). ; where L = upper limit(chain on table) and x = lower limit (chain that is hanging)
Subbed the givens and then solved.. where did I go wrong?

>>5
Equilibrium position is when dU/dr = 0.

>>6
Thanks. Now I feel stupid for not realizing that no force = equilibrium. Now I know I must have missed something in question 2 then.