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...

 

[rwisz]

As a scientist, it is important to approach problems like these with a clear understanding of the concepts involved. In this case, the problem involves the concepts of work and energy. Work is defined as the force applied over a distance, while energy is the ability to do work. In this problem, the two masses are initially at rest, meaning they have no kinetic energy. As they fall and rise, work is being done on them by the force of gravity, which is converting potential energy into kinetic energy. This can be represented by the equation:

W = ΔE = ΔK + ΔU

Where W is the work done, ΔE is the change in energy, ΔK is the change in kinetic energy, and ΔU is the change in potential energy.

We can also use the conservation of energy principle, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, the initial potential energy of the system is equal to the final kinetic energy of the masses before m2 hits the ground. This can be represented by the equation:

Ei = Ef

m1gh = (1/2)m1v1^2 + (1/2)m2v2^2

Solving for v2, we get:

v2 = √(2gh)

Plugging in the given values, we get:

v2 = √(2*9.8*1.2) = 4.85 m/s

This is the speed of the masses just before m2 hits the ground. It is important to note that this is the final speed of the system, as all of the potential energy has been converted into kinetic energy.