Application of Newton's Second Law: Weighted chain 
Q. A 750.0-kg boulder is raised from a quarry 125 m deep by a long uniform chain having a mass of 575 kg. This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking.
(a) What is the maximum acceleration the boulder can have and still get out of the quarry, and
(b) how long does it take to be lifted out at maximum acceleration if it started from rest?

The way I approached this problem is to consider the net force on the boulder. The forces acting on the boulder is its weight and the tension of the chain at the bottom. The top of the chain is 2.5 weight of chain while at bottom it is 1.5 weight of chain which is equal to 8452.5 N and the weight of boulder is 7350N. I subtract 7350 from 8452.5 and divide it by mass of the boulder to get 1.47 m/s^2.
The approach of the solution is to treat the boulder and chain as composite bodies and the end result is different. What is the error in my approach?
I would appreciate any insight
 A: If
Mass of boulder: $\space\space m_b = 750kg$
Mass of chain:  $\space\space m_c =  575kg$
The maximum force that the chain can withstand , and the weight of the whole system respectively should be
$F_{max} = 2.5*m_c*g = (2.5)(575)(9.81) = 14101.88N$
$F_{net} = (m_b+m_c)g = (750+575)(9.81) = 12998.25N$
The difference between those forces will define the maximum acceleration without breaking the chain. Think of it like this, the combined mass of the chain and the boulder is what is pulling down ($F_{net})$, and pulling up can handle a force of ($F_{max}$), so the difference between them $(F_{max}-F_{net})$ is what can go towards the upwards motion of the system.
Then we just go back to Newton's second law using $F_{max}-F_{net}$ and find our acceleration
$F_{max}-F_{net} = (m_b+m_c)a$
$14101.88 - 12998.25 = (750+575)a$
$1103.63 = 1325a$
$a = \frac{1103.63N}{1325kg} = 0.83m/s^2$
Then we can find our time using $d = \frac{1}{2}at^2$ for zero initial velocity, $d = 125m$ and $a = 0.83m/s^2$
$t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2(125)}{0.83}} = \sqrt{\frac{250}{0.83}} = 17.36s$
