# Conservation of energy theorem problem

1. Nov 10, 2008

### fishingspree2

1. The problem statement, all variables and given/known data
2 boxes are connected by a massless rope. the system is frictionless
m1 = 3kg
m2 = 5kg
http://img388.imageshack.us/img388/2033/problemwr4.jpg [Broken]
Find m1's speed when it is 1.8 m above the ground.
2. Relevant equations
K = 0.5mv^2
U = mg delta y

3. 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,
can anyone help me find what is wrong in my reasoning?

Last edited by a moderator: May 3, 2017
2. Nov 10, 2008

### borgwal

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

Otherwise, you did the right thing.

3. Nov 11, 2008

### Coto

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

$$v_1 = -v_2$$

4. Nov 11, 2008

### fishingspree2

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

5. Nov 11, 2008

### 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.

6. Nov 11, 2008

### 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.

Last edited: Nov 11, 2008