Conservation of energy theorem problem

[fishingspree2]

Homework Statement

2 boxes are connected by a massless rope. the system is frictionless
m₁ = 3kg
m₂ = 5kg
http://img388.imageshack.us/img388/2033/problemwr4.jpg
Find m₁'s speed when it is 1.8 m above the ground.

Homework Equations

K = 0.5mv^2
U = mg delta y

The Attempt at a Solution

v₁=v₂ obviously
also, delta y of m₁ = 1.8, whereas delta y of m₂=-1.8
At the beginning:
Etot = K₁ + K₂ + U₁ + U₂
since K₁, K₂ and U₁ are = 0, Etot = U₂ = m₂*g*5 = 25g

When m₁ is 1.8 meters above the ground:
Etot = K₁+K₂+U₁+U₂
25g = 0.5m₁v²+0.5m₂v²+1.8m₁g-1.8m₂g

solving for v gives 8.37,
but the answer is 2.97
can anyone help me find what is wrong in my reasoning?

 

[borgwal]

m2 ends up not at -1.8m, but at (5-1.8)m.

Otherwise, you did the right thing.

 

[Coto]

Also, be careful about your velocities. Though it doesn't matter in this problem since velocities are squared, it should be:

<br /> v_1 = -v_2 <br />

 

[fishingspree2]

m2 ends up not at -1.8m, but at (5-1.8)m.

Otherwise, you did the right thing.

I thought that what is important is the difference of height, and not the height itself. since the formula is U = mg delta y.. then delta y = [(5-1.8] - 5] = -1.8
humm can you please tell me where I am mistaken?

EDIT: I also see in the book U=mgh. If I had used this formula I woulnd't have had problems. What is the difference between both? When to use one instead of the other?

thank you

 

[borgwal]

You did NOT use the change in height: you wrote down the initial potential energy as mgh_2, with h_2=5m, then in the final potential energy you used h_2=-1.8m. That's inconsistent. Either you use the change in kinetic energy, or you use the correct initial and final potential energies.

 

[Coto]

You have to set an absolute coordinate system and work from there. i.e. the ground is y=0.

The \Delta y is most likely referring to the fact that potential energy is always relative with respect to some reference point. If we set the ground to be y=0 then \Delta y is just y ( \Delta y = y - 0 = y.

On the other hand, if we set sea level to be 0 but then do experiments in a city not at sea level, well \Delta y = y - y_0 where y_0 is the height above sea level, and y is height with respect to sea level. (Though at this point, it would make more sense to set y_0 to be zero, and take y to be height above the ground you're using).

With this in mind, you can see that \Delta y
is almost always just y.