# Energy Problem

1. Nov 21, 2006

### BSCS

A 20.0 kg block is connected to a 30.0 kg
block by a string that passes over a light frictionless
pulley. The 30.0 kg block is connected to a spring that
has negligible mass and a force constant of 250 N/m,
as shown in the figure. The spring is unstretched
when the system is as shown in the figure, and the
incline is frictionless. The 20.0 kg block is pulled 20.0
cm down the incline (so that the 30.0 kg block is 40.0
cm above the floor) and released from rest. Find the
speed of each block when the 30.0 kg block is 20.0 cm
above the floor (that is, when the spring is
unstretched).

The solution says to take the lowest point reached on the ramp by the 20.0 kg block as Ug = 0.

Then, it takes Ug of the 30.0 kg block to be based on +20.0 cm... I don't understand this. Isn't the y-distance from the lowest point on the ramp of the 20.0 kg block to the highest point of the 30.0 kg block *not* 20.0 cm?

2. Nov 21, 2006

### ponjavic

That sounds very strange, could you show the solution or an answer and what your problem in solving the question is?

3. Nov 21, 2006

### BSCS

$$\Delta U_{g} = (m_{2}sin\theta-m_{1})gx$$

$$\Delta U_{g} = [20.0 kg sin40 - 30.0 kg](9.80)(0.200) = -33.6 N$$

4. Nov 21, 2006

### Dorothy Weglend

Why would this distance be of any interest? For Ug, all you would be interested in is the change in the height of each block.

So, for the block on the incline, that would be the change in vertical distance as it travels back up the incline, and for the hanging block, it would just be 20 cm.

Or am I nuts? :uhh:

Dorothy

5. Nov 21, 2006

### BSCS

I can't comment on your sanity , but that goes back to my problem... How can it be 20 cm for the hanging block if Ug is not 0 at the point we would expect?

6. Nov 21, 2006

### Dorothy Weglend

Well, the hanging block would lose graviational energy, as the block on the incline would gain gravatational energy...

You have the solution, I guess? I get 1.24 m/s... Is that right?

(Just trying to confirm my sanity, I guess)

Dorothy

7. Nov 21, 2006

### ponjavic

The blocks have different Ug points. As what matters, as Dorothy said, is the difference in height between the two states. Having one common Ug point would make matters more complicated. So the Ug point for the first block is yes at the lowest point but for the other block it is at the top of the spring thus it will become 20cm.

8. Nov 21, 2006

### BSCS

1.24 m/s is correct.

9. Nov 21, 2006

### BSCS

Ok, this makes sense. So, I can have a separate Ug = 0 point for each element in the system?

10. Nov 21, 2006

### ponjavic

Indeed, it is just a matter of definition.

As all you are using is the difference in height, thus your definition of the ug point does not matter. Say your Ug point was on mars and then the block was raised by 20cm.

difference in potential energy: mgh2-mgh1=mg(h2-h1)=mg(dmars+20-dmars)=mg*20

11. Nov 21, 2006

### Dorothy Weglend

For me, the key to understanding these things it to think of the system energy. In the present case, the 20 kg block will have zero gravitational energy, but the 30 kg block will not, it will have some Ug.

When the system moves, then the 20kg block will gain Ug, as the 30 kg block will lose it. In other words, for the final state of the system, you will have a minus sign in front of the term with the 30 kg in it.