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  • what is internal and external force? How do I decide what forces are internal and external?

  • in work-energy theorem and conservation of mechanical energy, do we consider internal, external, or both forces?

Please explain with an example because at this point everything I learned about energy feels like it's in a blender that is switched on

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4 Answers 4

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First you have to decide what is and isn't part of your "system".

Let's say I have a bunch of masses orbiting each other. These will exert forces on each other - internal forces - and total energy and momentum will both be conserved.

Now let's imagine I reach in and poke one of the masses. This is an external force. Their total energy and momentum will change as a result of my interference, so it appears energy conservation has been violated. But in fact, by Newton's 3rd Law, we know that I experienced an equal and opposite force, and so I lost as much momentum as the system gained. Any energy I added to the system must also have been lost by me (we're ignoring friction). So if I'm included as part of that system, everything is still conserved.

I'm not sure of the exact context of your question but perhaps these "external forces" could be if you're considering a moving reference frame, like say inside a car going round a corner - then you experience G-forces, which appear to violate conservation of energy and momentum only because of your reference frame. Or gravity, for example, where we ignore the energy and momentum of the Earth - but we typically make up for that by considering the energy to have come from/gone into a gravitational potential.

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  • $\begingroup$ So when there are internal forces, energy and momentum is conserved... Are there any other properties like that? $\endgroup$
    – android
    Commented Jun 3 at 10:41
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You define a system.
Any internal forces within the system will be Newton Third Law pairs, ie for every internal force on part of system A due to part of system B there will be another internal force on part of system B due to part of system A which will have the same magnitude and is opposite in direction.

The problem with your second question is that the work-energy theorem is not defined the same by all Physicists.
One definition is that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle and this definition can be extended to rigid bodies.
The system is the particle / rigid body and all the forces are external.

As an example consider a rigid block of mass $m$ in a uniform gravitational field strength $g$.
The rigid block is the system.
The gravitational force acting on the block $F_{gravitational} = mg$ is an external force.
The block is allowed to fall a distance $h$.
The work done by the external force is $F_{gravitational} \times h = mgh$ and this is equal to the gain in kinetic energy of the block.
Note that of the system considered, the block, $mgh$ is not the loss of potential energy of the system, rather it is the work done by the eternal force.

Now consider the block, mass $m$, and the Earth, mass $M$,producing the gravitational field as the system.
The force on the block due to the Earth is an internal force and the other internal force is the force on the Earth due to the block, equal in magnitude and opposite in direction.
There are no external forces acting on the system.
The two internal forces will do work as the block is falling towards Earth and the Earth is rising up towards the block.
The work done on the block is $mgh$ where $h$ is the distance the block moves relative to the centre of mass of the system and this is equal to the gain in the kinetic energy of the block.
The work done on the Earth is $mgh'$ where $h'$ is the distance the block moves relative to the centre of mass of the system and this is equal to the gain in the kinetic energy of the Earth.
In most example the movement of the Earth and the work done on it is ignored because $h\gg h'$.

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  • $\begingroup$ Excellent points! Up and coming physicists would do well to harken unto the words in this post. The points discussed herein are subtle yet important. $\endgroup$ Commented Jun 3 at 19:45
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What is internal and external force ? How do I decide what forces are internal and external ?

An internal force is a force that one component of a system exerts on another. An external force is a force exerted on a system from outside.

Note that this means that whether a force is internal or external depends not on the nature of the force itself, but on where you put the boundaries of your system. Consider a block sitting on the ground. If you decide that your system is just the block, then the block's weight is an external force. However, if you decide that your system is the block and the Earth then the block's weight is an internal force.

In work-energy theorem and conservation of mechanical energy, do we consider internal, external, or both forces ?

Energy is always conserved. However, only an external force can do work on a system and so change its total energy. Internal forces - as long as they are conservative - do not change the total energy of a system (although they may exchange potential energy for kinetic energy or vice versa) so we often ignore them.

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    $\begingroup$ "Internal forces cannot change the total energy of a system" This is completely wrong. Internal forces cannot change the momentum, but they can very well increase or decrease the mechanical energy. Consider the simple case of an inelastic collision between two blocks. $\endgroup$ Commented Jun 3 at 16:40
  • $\begingroup$ The total energy in the system hasn't changed. It's just been converted to heat energy. But you're right in that it's important to consider that frictional forces can cause "loss" of energy in this way. $\endgroup$
    – JaredJeya
    Commented Jun 19 at 8:41
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Internal forces are those forces which originate from the the system itself and are exerted by the system on its surroundings.These forces may also be exerted by components of the system on other componets present within the system.

For Ex consider a block of mass M present on the frictionless upper surface of a table. Here we are considering the block as our system. Image 1

Here the block is applying a force called normal force on the table. (Not to be confused by gravitational force being exerted on the block by the Earth) Since this force is being applied By the block on the table it would be counted as an internal force(The force is being generated as the result of interaction of the atoms of the lower most surface of the block and the uppermost surface of the table.)

The electrostatic or gravitational attraction and repulsion within the block can also be considered as internal forces since these are being generated within the system(ie block)itself.

Now coming to external forces,these are those forces which are being applied On the system by surroundings.

Taking the same example:

Image 2

In this case the forces are being applied by the block by the surrounding. A vector represents the gravitational force which is being applied by the earth on the block whereas B vector represents the normal force which is being applied by the table on the block(Not by the block on the table as discussed in internal forces.)

If we consider the table and the block as our system both the normal reactions would now be considered as internal forces. However the gravitational force would still be considered as an external force since it is being applied by the earth which will act as a surrounding for our new system.

Hope this clear your doubts.

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