Homework Help: 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