Conservation of Mechanical Energy - Which Equation To Use?

[chaneth8]

Homework Statement

Finding The Maximum Height Reached By A Cube

Relevant Equations

Change in mechanical energy of a system = change in kinetic energy + change in potential energy

Hi, I'm working on part a of the following mechanics problem from MIT Open Courseware: https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/resources/mit8_01f16_pset9/:

I think the key to solving this problem relies on the conservation of mechanical energy. Since there are no nonconservative forces acting on this system, total mechanical energy is conserved.

However, it's not immediately apparent to me what I should consider the system as. I can treat the cube as my system, or the block and the cube together as my system. Both seem equally valid to me. All the forces acting on the cube are conservative (the normal force on the block is perpendicular to the direction it's travelling in, so we need not consider it). Similarly, all the forces acting on the block are conservative. So conservation of mechanical energy should hold for both.

Before we proceed, let us denote the final velocities of the cube and the block as $v_\rm{final}$.

If we consider only the cube as our system when using conservation of mechanical energy, we get the following equations:
By conservation of mechanical energy:
$\frac 1 2 m_\rm{c} (v_\rm{c,0})^2 = mgh + \frac 1 2 m_\rm{c} (v_\rm{final})^2$

Assuming the time it takes for the cube to climb up the block is small, we can equate the momentum at the initial and final times:
$m_\rm{c} v_\rm{c,0} = (m_\rm{c} + m_\rm{b}) v_\rm{final}$

However, if we consider the cube and the block as our system when using conservation of mechanical energy, we get the following equations:
By conservation of mechanical energy:
$\frac 1 2 m_\rm{c} (v_\rm{c,0})^2 = mgh + \frac 1 2 m_\rm{c} (v_\rm{final})^2 + \frac 1 2 m_\rm{b} (v_\rm{final})^2$

Assuming the time it takes for the cube to climb up the block is small, we can equate the momentum at the initial and final times:
$m_\rm{c} v_\rm{c,0} = (m_\rm{c} + m_\rm{b}) v_\rm{final}$

These are two sets of equations - they would give us different results. So how do I determine which set is valid?

 

[jbriggs444]

However, it's not immediately apparent to me what I should consider the system as. I can treat the cube as my system, or the block and the cube together as my system. Both seem equally valid to me. All the forces acting on the cube are conservative (the normal force on the block is perpendicular to the direction it's travelling in,

Think about this last claim for a moment.

1. If you are taking the cube as your system then the normal force of cube on block is irrelevant. You need to be considering the normal force of block on cube.

2. Is the normal force of block on cube really perpendicular to the direction the cube is travelling?

Similarly, all the forces acting on the block are conservative. So conservation of mechanical energy should hold for both.

So the block does not retreat under the normal force from the cube? It remains motionless despite being subject to an unbalanced net force?

 

[chaneth8]

Think about this last claim for a moment.

1. If you are taking the cube as your system then the normal force of cube on block is irrelevant. You need to be considering the normal force of block on cube.

2. Is the normal force of block on cube really perpendicular to the direction the cube is travelling?

Correct me if I'm wrong, but isn't the normal force of the block on the cube perpendicular to the direction the cube is travelling? The block is circular, so the cube is travelling tangent to the surface of the block. Meanwhile, the normal force of the block on the cube acts in the radial direction, so the normal force is perpendicular to the direction the block is moving.

So the block does not retreat under the normal force from the cube? It remains motionless despite being subject to an unbalanced net force?

I see my mistake now - there is a force pushing the block forward in the same direction as it has motion, so I can't assume that ignore the work done by the normal forces on the block.

 

[jbriggs444]

Correct me if I'm wrong, but isn't the normal force of the block on the cube perpendicular to the direction the cube is travelling?

No. It is not.

It would be perpendicular if the block were not moving.

I see my mistake now - there is a force pushing the block forward in the same direction as it has motion, so I can't assume that I can ignore the work done by the normal forces on the block.

As you seem to have realized, the block is moving.

The work done by the normal force of cube on block will be equal and opposite to the work done by the normal force of block on cube. Neither will be zero.

It is most convenient to define the system boundaries so that this force pair is internal and may be ignored. In the absence of friction, internal contact force pairs do not drain mechanical energy.

 

[chaneth8]

No. It is not.

It would be perpendicular if the block were not moving.

Could you elaborate a bit more on this part? I thought when an object moves in a circle, the direction of motion is always tangent to the circle - and we are given that the ramp is circular.

Why would the block being in motion change that?

It is most convenient to define the system boundaries so that this force pair is internal and may be ignored. In the absence of friction, internal contact force pairs do not drain mechanical energy.

I see - so we consider the block and the cube as one system, so that the work done by their internal forces cancel out - is that right?

 

[jbriggs444]

Could you elaborate a bit more on this part? I thought when an object moves in a circle, the direction of motion is always tangent to the circle - and we are given that the ramp is circular.

Why would the block being in motion change that?

If you superimpose a linear motion on a circular motion, you get a cycloid. Like the path drawn out by a point on a tire.

If you superimpose a non-uniform linear motion on a non-uniform circular motion as in the case at hand, the resulting motion will be more difficult to characterize. Fortunately, we can invoke conservation laws and ignore the path from initial to final state. As long as the forces along the path are energy-conserving, all is well.

I see - so we consider the block and the cube as one system, so that the work done by their internal forces cancel out - is that right?

Yes, we had a thread on that recently.

There are some complexities depending on exactly what sort of work one is contemplating and exactly what sort of energy one is thinking of as being conserved or transferred.

"mechanical work" (thank you, @Dale) contemplates force times the [parallel] displacement of the material at the point where the force acts. The result tells you how much mechanical energy has been transferred.

This one is the one I have in mind. An internal normal force pair never changes the system mechanical energy. The normal components of the motion are equal. The forces are equal and opposite. The total of the two equal and opposite works is obviously zero.

By contrast an internal [kinetic] frictional force pair can obviously change system mechanical energy.

 

[kuruman]

These are two sets of equations - they would give us different results. So how do I determine which set is valid?

To answer your original question, you can be sure that the choice of cube + block system equations are valid because (a) the system's mechanical energy is conserved (no friction anywhere) and (b) the system's momentum in the horizontal direction is conserved (no horizontal external forces). The two equations that follow from these two considerations, if written correctly, must be valid.

 

[chaneth8]

An internal normal force pair never changes the system mechanical energy. The normal components of the motion are equal. The forces are equal and opposite. The total of the two equal and opposite works is obviously zero.

By contrast an internal [kinetic] frictional force pair can obviously change system mechanical energy.

I've had a think about this but there is still something I don't get - could you explain why a normal force pair never changes the system's mechanical energy? I understand that the forces within the pair are equal and opposite, but when we consider the system's mechanical energy, we need to consider the work done by these forces.

The work done on an object by a force is given by a line integral of the force, with respect to the path travelled by the object. So if the two objects travel in two different paths, even if the normal forces they exert on each other are equal and opposite, how can we be sure that these two works cancel out to 0?

 

[jbriggs444]

I've had a think about this but there is still something I don't get - could you explain why a normal force pair never changes the system's mechanical energy? I understand that the forces within the pair are equal and opposite, but when we consider the system's mechanical energy, we need to consider the work done by these forces.

The work done on an object by a force is given by a line integral of the force, with respect to the path travelled by the object. So if the two objects travel in two different paths, even if the normal forces they exert on each other are equal and opposite, how can we be sure that these two works cancel out to 0?

Do not bother with following the path or evaluating a path integral. Look at the differential.

How much incremental work is done by the normal component of an internal contact force pair over an infinitesimal time element, $dt$?

There are two force components in the pair. The normal force from A on B. And the normal force from B on A. You agree that these are equal and opposite as a consequence of Newton's third law, surely?

There are two incremental motions. One for object A and one for object B. There is no requirement that these be equal. However...

Let us split the incremental motions into components -- a normal component toward or away from the other object and a tangential component parallel to the interface.

The tangential components are obviously irrelevant. No force under consideration acts in the tangential direction. We are considering only the normal force. So we need consider only the normal component of the motion of each object at their shared interface.

Because it is a contact force, those two normal components are identical. The surfaces are staying together.

If we label the normal component of the incremental motion at the shared interface as $d y$ and the normal force of A on B as $F_n$ then the work done by A on B is $\vec{F_n} \cdot \vec{dy}$ while the work of B on A is $-\vec{F_n} \cdot \vec{dy}$. The two incremental works are equal and opposite. Their sum is zero

The sum of the path integrals is equal to the integral of the sum. The integral of zero is zero.

 

[kuruman]

$\dots$ but when we consider the system's mechanical energy, we need to consider the work done by these forces.

Why do we "need" to consider the internal forces? If by "system" you mean cube + block we need only consider the forces that cross the system's boundary and are external to both cube and block. The only such force is gravity that does work only on the part of the system called "cube". That's the only work that we have to calculate from the point where the cube makes first contact with the block to the point of maximum height where the two system components move horizontally as one.

The correct way to write the energy equation for the two-component system is the work-energy theorem: $$\begin{align} & \Delta K_{\text{cube}}+\Delta K_{\text{block}}=W_{\text{gravity}} \nonumber \\
& \left(\frac{1}{2}m_cv_{\text{final}}^2-\frac{1}{2}m_cv_{\text{c,0}}^2\right)+\left(\frac{1}{2}m_bv_{\text{final}}^2-0\right)=-m_cgh. \nonumber \\
\end{align} $$It is algebraically the same equation that you put down and shows where the terms come from. Note that the gravitational potential energy is not part of it. That's because the Earth is not part of the system by choice. Therefore, the Earth exerts an external force on the system and the work it does must be on the right-hand side of the equation.

I could say more, but I will stop here.