How to find the equilibrium position and potential energy?

[KFC]

Homework Statement

Given a setup with spring, the massless spring constant is k, unstretched length of the spring is L_0. A mass m is attaching to the top of the spring. We only know the height of the mass (measuring from the top of the platform) is x, the height of the platform is h. Try to find the equilibrium position of the system, the potential energy and kinetic energy of the system in terms of x.

http://img150.imageshack.us/img150/5247/a101pc3.jpg

2. The attempt at a solution
I setup a coordinate system with the origin on CD line, call the vertical axis X (capital X). The hooke law gives

mg = -k\Delta X, \qquad \Delta X = x + h - L_0

For finding the equilibrium position (x_0), we let \Delta X = x + h - L_0 = 0 gives x_0 = L_0 - h

The kinetic energy of the mass is obvious

E_k = \frac{m}{2}\left(\frac{dX}{dt}\right)^2 = \frac{m}{2}\left[\frac{d(x+h)}{dt}\right]^2 = \frac{m}{2}\left(\frac{dx}{dt}\right)^2

For the potential energy, we have
E_p = \frac{k}{2}(\Delta X)^2 = \frac{k}{2}\left(x + h - L_0)^2

But the answer for the potential energy is
E_p = \frac{k}{2}x^2

How is that possible?

 

[chaoseverlasting]

There are a couple of things which I am not too sure about. Is it given that x+h is the equilibrium position of the system (ie the system is in equilibrium in the diagram?). If it isnt, then your equation \Delta x=x+h-L_0 is not valid.

Even if the system is in equilibrium, why would \Delta x=x+h-L_0=0 give you the equilibrium position? That would give you the initial position of the spring when its uncompressed.

 

[KFC]

Thanks for reply. The problem doesn't tell x+h is the equilibrium position. Well, for vertical spring, I think the equilibrium position is where the force (or acceleration) is ZERO. So I let mg=0 that gives \Delta X=0. Am I doing anything wrong? If so, please help me. Thanks

There are a couple of things which I am not too sure about. Is it given that x+h is the equilibrium position of the system (ie the system is in equilibrium in the diagram?). If it isnt, then your equation \Delta x=x+h-L_0 is not valid.

Even if the system is in equilibrium, why would \Delta x=x+h-L_0=0 give you the equilibrium position? That would give you the initial position of the spring when its uncompressed.

 

[chaoseverlasting]

If you assume cd to be the origin line, then let the distance of the block from the line be y. As such, the potential energy of the block is given by mgy.

Also, the kinetic energy of the block is \frac{1}{2}mv^2 where v=\frac{dy}{dt}.

Now you have to express v² in terms of y to get the expression for kinetic energy.

To do this you need express find the force equation of the block in terms of y. This gives you F=mg-k(x+h-y). From this you can get the value of \frac{v^2}{2} and hence the expression for kinetic energy.

Now you know the expressions for KE and PE. Add them up and you get total energy. At equilibrium, the total energy must be minimum. You can find the minima of the system as the expression for the total energy will be a function of y, which will give you the equilibrium position.

I know its a little long, maybe someone else can come up with a shorter way to do it.

 

[chaoseverlasting]

Thanks for reply. The problem doesn't tell x+h is the equilibrium position. Well, for vertical spring, I think the equilibrium position is where the force (or acceleration) is ZERO. So I let mg=0 that gives \Delta X=0. Am I doing anything wrong? If so, please help me. Thanks

Youre right, the force must be zero, but that is the NET force acting on the block. In this case, the net force is the gravitational force MINUS the force of the spring. That value must be zero, not mg.

 

[KFC]

Ok, I just update the figure and add the coordinate system. With your hints, I reconsider the solution and note that the problem require to express all answers in terms of x.

1) First calculate the kinetic energy, I know that (as you told) the the kinetic energy will be

E_k = \frac{m}{2}\left(\frac{dy}{dt}\right)^2

Since y=x+h, so we also have

E_k = \frac{m}{2}\left(\frac{dx}{dt}\right)^2

2) For finding the equilibrium position, we must set the total force to zero, let the unstretched length of spring be L_0

F = mg - k(x+h - L_0) = 0

gives the equilibrium position (measure from the top of the platform) be

x_0 = \frac{mg}{k}(L_0 - h)

3) The elastic potential energy in terms of x is
E_{ep} = \frac{k}{2}(x - x_0)^2


Hope these results are correct :)

If you assume cd to be the origin line, then let the distance of the block from the line be y. As such, the potential energy of the block is given by mgy.

Also, the kinetic energy of the block is \frac{1}{2}mv^2 where v=\frac{dy}{dt}.

Now you have to express v² in terms of y to get the expression for kinetic energy.

To do this you need express find the force equation of the block in terms of y. This gives you F=mg-k(x+h-y). From this you can get the value of \frac{v^2}{2} and hence the expression for kinetic energy.

Now you know the expressions for KE and PE. Add them up and you get total energy. At equilibrium, the total energy must be minimum. You can find the minima of the system as the expression for the total energy will be a function of y, which will give you the equilibrium position.

I know its a little long, maybe someone else can come up with a shorter way to do it.