Question regarding gravitational force as external force

Question

So, I was watching a lecture on YouTube for problems on conservation of energy and momentum and I don't quite understand this:

In this question, mass $M$ is released from the peak of the smooth movable wedge kept on a smooth horizontal surface and we are required to find speed of both when block reaches ground. Now, I do understand for mass+wedge system, along horizontal momentum is conserved as their are are no external horizontal forces but along vertical their is a vertical external force that is gravitational force. So we can apply momentum conservation along horizontal and not in vertical because of gravitational force.

My doubt is if gravitational force is considered as an external force this mean it changes the mechanical energy of the system so how come can we apply Total mechanical energy conservation?

As far as my understanding (that may be wrong) for conservative forces and law of conservation of total mechanical energy, for conservation of total mechanical energy there shouldn't be any work due to external forces, not even conservative forces that are external so that would mean their is no gravitational potential energy and only work done by gravity. I don't understand this are all conservative forces have potential energy irrespective of being internal and external?

Answer 1

My doubt is if gravitational force is considered as an external force

Yes, it is considered as an external force, an external conservative force.

...so how come can we apply Total mechanical energy conservation?

Because the force is a conservative force, meaning that it results from a change in potential energy: $$ \vec F_{grav} = -\vec \nabla U_{grav} $$

As far as my understanding(that may be wrong) for conservative forces and law of conservation of total mechanical energy, for conservation of total mechanical energy there shouldn't be any work due to external forces,

The work done by the net external force changes the kinetic energy. This is the content of the work-kinetic-energy theorem.

I don't understand this are all conservative forces have potential energy irrespective of being internal and external?

The work done by the external gravitational force in this case is: $$ \int d\vec x \cdot \vec F_g = -\int d\vec x \cdot \vec \nabla U_g = -\Delta U_g $$

The work done by the net external force is: $$ \int d\vec x \cdot \vec F_{net} = +\Delta K $$

In your case it looks like the net external force and the external gravitational force are one and the same, thus: $$ \Delta K = -\Delta U_g\;. $$

Or, $$ \Delta K + \Delta U_g = 0 $$

Or. $$ \Delta E = 0\;, $$ where $E = K + U_g$.

If there are other conservative forces at play, one still arrives at: $$ \Delta E = 0 $$ as long as you include all the conservative force potentials in the definition of total mechanical energy. I.e., $$ E = K + U\;, $$ where $U$ is the sum of all the conservative force potential energies.

Answer 2

If you chose the system to be the mass and the wedge then the gravitational force due to the attraction of the Earth is an external force to your system and you should equate the work done on the system by the gravitational force and the change in kinetic energy of the system.

If you want to use the concept of gravitational potential energy then your system must include the Earth.
In that case the change in gravitational potential energy plus the change in kinetic energy should be equal to zero, ie mechanical energy is conserved as there are no external forces acting on the mass, wedge and Earth system.

Answer 3

I will use Lagrangian method for the simple reason that the system is complex so, and we need to understand better. The small ones stand for the object and the capitals for the ramp. The Lagrangian is

$$L=\frac{M\dot{X}^2}{2}+\frac{m((\dot{X}+\dot{x})^2+\dot{y}^2)}{2}-MgY-mgy$$.

I have some equations that have to be fulfilled: $h_1=Y=0$ and $h_2=y+ax=0$.

I assume the ramp is not curvy with sloop -1. Also the x,y axis are non-static. By replacing the bonds the "information" of the Lagrangian remains intact and becomes:

$$L=\frac{M\dot{X}^2}{2}+\frac{m((\dot{X}+\dot{x})^2+a^2\dot{x}^2)}{2}+mgax$$

We know that $P=\frac{\partial L}{\partial \dot{X}}$ and $p=\frac{\partial L}{\partial \dot{x}}$

So the Jacobi integral gives: $$E=p\dot{x}+P\dot{X}-L$$ Do the $\frac{dE}{dt}$, it should be 0(if it is not i did something wrong). The E refers to mechanical energy.