How Does Atwood's Machine Demonstrate Conservation of Energy?

[LastXdeth]

Homework Statement

The two masses in the Atwood’s machine shown in Figure 8–23 are initially at rest at the same height. After they are released, the large mass, m2 falls through a height h and hits the floor, and the small mass, m1rises through a height h. Find the speed of the masses just before m2 lands if h = 1.2m, m1 = 3.7kg and m2 = 4.1kg

Homework Equations

E=Eo
Ug=mgh
K=(1/2)mv^2

The Attempt at a Solution

m1*g*y +m2*g*y = (1/2)*m1*v^2 + (1/2)*m2*v^2

I factored outed g*y and (1/2)*m*v^2
g*y*(m1 + m2) = (1/2)*(v^2)*(m1 + m2)

I canceled out (m1 + m2)
g*y = (1/2)*v^2

g*y = (1/2)*v^2

2*g*y = v^2

√(2*g*y) = v

√(2*9.8*1.2) = v

This is apparently wrong:
4.85m/s ≈ v

 

[issacnewton]

first define the reference level for the potential energy... what level have you chosen ?

 

[LastXdeth]

I would like to set the reference level at where the dotted line is (where v=0).

 

[issacnewton]

ok, so that means your initial total potential energy is zero. Also initial total kinetic energy
is zero. So total initial energy is zero. What about the final configuration ? The left block has climbed up distance h , so its potential energy is $m_1gh$ and the second
block is gone below the reference level. so its potential energy would be
$-m_2gh$. What about their kinetic energies ? Find that expression and use the conservation of energy theorem, which implies that the total initial energy must be equal to the total final energy...