Message No.4186 from drg , 28 September 2008 21:08:
 îòâåò íà No.3741: HELP!!! Masses on a pulley îò Jake , 10 April 2008:
Ok, the answer might help someone else.
First get the acceleration using the equation (v2)= 2a(x-xo)
Since the initial velocity is 0 and the acc is constant.
So a=(2*2)/(2*1.4)=2/1.4 m/s2
Use Newton's Second Law
∑Fext= (m1+m2)a
You know that m1+m2=13
so Mtotal*a=13*(2/1.4)
but ∑Fext= m2g-m1g
if m2 is the larger block, which is pulling down.
Two unknowns, two equations.
(1) m1+m2=13
(2) g(m2-m1)=13*(2/1.4)
Solve it and you'll get m2=7.45 rounded up ;)