Spring potential Problem Reference Point

[yolo123]

Hi, I uploaded the problem and the solution's manual answer.
There is something wrong in it I think.

For this problem:
"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. 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)."

I think that the part in bold is wrong. The 30 kg mass is heavier than the 20 kg mass, so the pulley will spin clockwise, thereby compressing the spring and giving it a new equilibrium point. Would this change the answer to the problem? The initial and final states will differ because of the spring being stretched.

 

[Doc Al]

I think that the part in bold is wrong. The 30 kg mass is heavier than the 20 kg mass, so the pulley will spin clockwise, thereby compressing the spring and giving it a new equilibrium point. Would this change the answer to the problem? The initial and final states will differ because of the spring being stretched.

Why would the equilibrium position of the system matter?

 

[yolo123]

If I set up the equation using the "new" equilibrium point:
Let h be the height difference between the 20cm from ground and the "true?" equilibrium point.

30(h+20)g+1/2(k)(h+20)^2=20(9.8)sin(40)(0.2)+1/2(20)v^2+1/2(30)v^2+1/2(250)h^2

This is different than what it is on the solution's manual.

 

[yolo123]

So, let me get this straight.
You're trying to tell me that:
If I have a spring with a mass oscillating vertically with an amplitude of 10 cm (20 cm max displacement).

If I set my x=0(eq. point) to the lowest point that the mass goes, and I calculate the energy at the highest point:
I get 1/2k(20)^2+mg20

If I set my x=0 to 1 cm above lowest point:
I get 1/2k(19)^2+mg(19)

Now, I can accurately say that 1/2k(19)^2+mg(19)=1/2k(20)^2+mg20?

 

[Doc Al]

If I set up the equation using the "new" equilibrium point:
Let h be the height difference between the 20cm from ground and the "true?" equilibrium point.

30(h+20)g+1/2(k)(h+20)^2=20(9.8)sin(40)(0.2)+1/2(20)v^2+1/2(30)v^2+1/2(250)h^2

This is different than what it is on the solution's manual.

What matters is the energy change once the spring is released, which depends on the amount of stretch, not the displacement from some new equilibrium point.

The book's answer is correct.

 

[Doc Al]

So, let me get this straight.
You're trying to tell me that:
If I have a spring with a mass oscillating vertically with an amplitude of 10 cm (20 cm max displacement).

Who says that the mass is oscillating with an amplitude of 10 cm?

Forget about any oscillation (that would be another problem). You stretch the spring and let it go. When the spring is back to its original unstretched length, what's the speed of the masses? That's the problem.

To find the energy in the spring you measure the amount of stretch from its unstretched length.

 

[yolo123]

But the amount of energy change changes because you're using a quadratic equation, no?

 

[Doc Al]

But the amount of energy change changes because you're using a quadratic equation, no?

I don't know what you mean by that.

 

[yolo123]

If you look at this graph:
The amount of kinetic energy change is not equal from similar displacements.

 

[yolo123]

Ok I think I got this.

If I look at the change in energy for the spring with mass on from the EQUILIBRIUM POINT to x m of displacement from EQUILIBRIUM POINT.

If I look at a spring without the mass on and consider the new EQUILIBRIUM POINT, and consider x m of displacement from the new EQUILIBRIUM POINT, I will get the same energy change as the first case.Please confirm. I know I may be "slow".

 

[Doc Al]

Ok I think I got this.

If I look at the change in energy for the spring with mass on from the EQUILIBRIUM POINT to x m of displacement from EQUILIBRIUM POINT.

If I look at a spring without the mass on and consider the new EQUILIBRIUM POINT, and consider x m of displacement from the new EQUILIBRIUM POINT, I will get the same energy change as the first case.

I'll put it this way. Say the new equilibrium point is a distance h below the unstretched position. When the spring is brought to the point where it is released, a distance of .2+h above equilibrium, what is the stored spring potential energy? Still only 1/2k(.2)^2.

 

[yolo123]

Is this not misleading: "The spring is unstretched when the system is as shown in the figure"?

How could you consider it unstretched if there is a mass on it? The appropriate way to formulate it would be the new equilibrium point is 20 cm above ground. Do you agree?

 

[yolo123]

If the spring is trully unstretched, the figure makes no sense because that would mean there is a force already pulling on the 20kg mass. Do you agree?

 

[Doc Al]

Is this not misleading: "The spring is unstretched when the system is as shown in the figure"?

No, it seems clear to me. It's a statement about the spring. (Make no assumptions about how the system got to such a position or whether the masses are moving or not.)

How could you consider it unstretched if there is a mass on it? The appropriate way to formulate it would be the new equilibrium point is 20 cm above ground. Do you agree?

No. You are assuming that the configuration shown is one where the masses are in equilibrium. It's not.

 

[Doc Al]

If the spring is trully unstretched, the figure makes no sense because that would mean there is a force already pulling on the 20kg mass. Do you agree?

Again, do not assume that the system shown is in equilibrium. (If it were, then you'd be right--an additional force must be supporting the mass.)