Free inclined plane and a block sliding on it

[xxxyyy]

Hi,
I'm missing something really stupid here...
The problem is the usual one with a block sliding down (or moving up, it should be the same) a frictionless inclined plane,which itself is free to move on a orizontal frictionless surface.
These problems are usually solved stating that only gravitation (conservative) does work.
Don't the reactions do work too?
The reaction on the block is orthogonal to the plane, but not orthogonal to the path of the moving block, the inclined plane is moving, so it does work (negative)
The reaction on the inclined plane has a component in the direction of the movement of the plane itself, so it does positive work.
They should cancel, but how to show it? Energy conservation because there's no friction? But if so they should have started from energy conservation, I guess?
Thank you!

 

[PeroK]

Hi,
I'm missing something really stupid here...
The problem is the usual one with a block sliding down (or moving up, it should be the same) a frictionless inclined plane,which itself is free to move on a orizontal frictionless surface.
These problems are usually solved stating that only gravitation (conservative) does work.
Don't the reactions do work too?
The reaction on the block is orthogonal to the plane, but not orthogonal to the path of the moving block, the inclined plane is moving, so it does work (negative)
The reaction on the inclined plane has a component in the direction of the movement of the plane itself, so it does positive work.
They should cancel, but how to show it? Energy conservation because there's no friction? But if so they should have started from energy conservation, I guess?
Thank you!

The tricky thing if you do this using forces is that you have an accelerating reference frame for the larger block.

 

[tech99]

Am I being naiive here? At the instant of letting go, we have just static forces. Only the horizontal ones create horizontal movement and these are equal to W x cos (slope).

Hi,
I'm missing something really stupid here...
The problem is the usual one with a block sliding down (or moving up, it should be the same) a frictionless inclined plane,which itself is free to move on a orizontal frictionless surface.
These problems are usually solved stating that only gravitation (conservative) does work.
Don't the reactions do work too?
The reaction on the block is orthogonal to the plane, but not orthogonal to the path of the moving block, the inclined plane is moving, so it does work (negative)
The reaction on the inclined plane has a component in the direction of the movement of the plane itself, so it does positive work.
They should cancel, but how to show it? Energy conservation because there's no friction? But if so they should have started from energy conservation, I guess?
Thank you!

As far as I can see, the PE of the block is converted to KE shared between the block and the plane. The horizontal forces are equal, and the momentum of the two is equal and opposite, but the energy is shared in the inverse ratio of the masses.

 

[DrStupid]

Don't the reactions do work too?

Yes, they do. Gravity does work on the block and the block does work on the plane. However, the work done by the reaction forces is just transfer of energy between block and plane and therefore doesn't change the total energy. That means you can start from energy conservation if you want (but you don't need to).

 

[A.T.]

The reaction on the block is orthogonal to the plane, but not orthogonal to the path of the moving block, the inclined plane is moving, so it does work (negative)

The reaction on the inclined plane has a component in the direction of the movement of the plane itself, so it does positive work.

They should cancel, but how to show it? Energy conservation because there's no friction?

They should cancel, because of energy conservation. You can check it by computing their motion using Newton's Laws and from that the work done on them by the contact forces.

 

[Nugatory]

As far as I can see, the PE of the block is converted to KE shared between the block and the plane. The horizontal forces are equal, and the momentum of the two is equal and opposite, but the energy is shared in the inverse ratio of the masses.

The problem is a bit trickier than that because the wedge is subject to three forces (gravity, normal force from the surface underneath, normal force from the block) while the block is subject to only two (gravity, normal force from the block). The wedge and the block aren't moving in opposite directions.

 

[Nugatory]

These problems are usually solved stating that only gravitation (conservative) does work.
Don't the reactions do work too?

Some texts word this more carefully than others.
Gravity is the only force that does work on the total system composed of the block and the plane. The reaction forces acting between the internal components of the system do work, but this only affects how the energy is distrbuted internally.

 

[A.T.]

The wedge and the block aren't moving in opposite directions.

Horizontally they are, if the block is sliding down.

 

[tech99]

Horizontally they are, if the block is sliding down.

Agree. My apologies, I did not mean to say that!

 

[A.T.]

Agree. My apologies, I did not mean to say that!

You didn't say that. @Nugatory did. My apologies for misquoting you, fixed now.

 

[Nugatory]

Horizontally they are, if the block is sliding down.

Yes, so we can use conservation of momentum to relate the horizontal components of the respective velocities to the ratio of the masses, but the block has a non-zero vertical velocity component as well so the energies aren’t related by the same ratio.

 

[tech99]

The momentums are equal but the energies are not. This is because Kinetic Energy is dependent on velocity squared, so a large mass moving slower will have less energy than a small mass moving faster, assuming the same momentum.
It seems to me that the horizontal forces created by the block in contact with the ramp do not fluctuate as the block accelerates, but remain constant. These forces are less than the weight of the block, due to the angle of the ramp. In addition, there are vertical forces at work but these do not alter the result for horizontal motion. The block has kinetic energy due to its vertical motion when it reaches the bottom of the ramp, and this together with the kinetic energy given to the block and ramp in horizontal directions will equal the total PE of the block at the moment of release.

 

[A.T.]

The momentums are equal but the energies are not. This is because Kinetic Energy is dependent on velocity squared, so a large mass moving slower will have less energy than a small mass moving faster, assuming the same momentum.

That's not the point @Nugatory is making. The point is that unlike in a 1D case, the KE of the block depends on its vertical velocity too.

You have 3 velocity componnets:

VBx : horizontal block velocity
VBy : vertical block velocity
VRx : horizontal ramp velocity

The are related by 3 equations:

VBx & VRx: horizontal momentum conservation
VBx & VBy & VRx: in the frame of the ramp the block moves parallel to the incline
VBx & VBy & VRx: energy conservation

It seems to me that the horizontal forces created by the block in contact with the ramp do not fluctuate as the block accelerates, but remain constant.

What are you basing this assumption on?

 

[etotheipi]

Maybe easier to just use variational principles. Let $q_1$ be the distance of the block from the bottom corner of the wedge of internal angle $\alpha$, and $q_2$ the distance moved by the corner of the wedge from initial point $\mathcal{O}$. You have for the block$$v^2 = (\dot{q}_1 \cos{\alpha} + \dot{q}_2 )^2 + (\dot{q}_1 \sin{\alpha})^2 = {\dot{q}_1}^2 + {\dot{q}_2}^2 + 2\dot{q}_1 \dot{q}_2 \cos{\alpha}$$Then for the system with these coordinates $q = (q_1, q_2)$$$\mathcal{L}(q, \dot{q}) \equiv \frac{1}{2} m ({\dot{q}_1}^2 + {\dot{q}_2}^2 + 2\dot{q}_1 \dot{q}_2 \cos{\alpha}) + \frac{1}{2} M {\dot{q}_2}^2 - mgq_1 \sin{\alpha}$$Now you obtain two coupled second order differential equations. Also note that $q_2$ is an ignorable coordinate, which in this case corresponds to the conservation of the projection of the momentum onto a horizontal basis vector, $m\ddot{q}_2 + m\ddot{q}_1 \cos{\alpha} + M\ddot{q}_2 = 0$.