1
$\begingroup$

I have seen many problems in physics where the solution starts wih "using conservation of energy, we get" without telling anything about why do we start with like this,

My concern is how do we know that we should use Conservation of Energy in a question ?

Is it like when we have to calculate velocity, distance, force we should use this law ? Or when there is motion of objects?

is it when time is not asked?

What could be the tag words for the same as otherwise solving physics problems becomes a guess work and goes into a rote study like history and geography that you have to solve the same problem previously to solve it which is not possible.

Improve this question
$\endgroup$
2
  • $\begingroup$ For an example, see this question: physics.stackexchange.com/q/712500. There are two answers: One answer calculates forces and involves cosines; the other gets the same result calculating energies with (I believe) much simpler mathematics. $\endgroup$
    – rghome
    Commented Aug 24, 2022 at 13:34
  • $\begingroup$ There are basically 3 things conserved in Newtonian Mechanics: Mass, Momentum, Energy. For any new problem, these are usually the big three to assess and see if they help relate the variables you know to the ones you don't know. Also geometric constraints, e.g. a piece is a rigid body, so dist between points on it remains constant. $\endgroup$
    – RC_23
    Commented Aug 24, 2022 at 21:33

5 Answers 5

Reset to default
2
$\begingroup$

The Conservation of Energy holds almost all the times but the thing is we can't use it to calculate "what portion of the energy is distributed in various forms ?" e.g. If you push a block resting on a table , the block moves and then stops after some distance.

Now if you touch the block, you can feel that it's hot. Also while it was moving you might heard some rubbing sound.

So basically using the Conservation of Energy you can tell that the whole initial kinetic energy is lost in other forms (heat, sound, electrical sparks) but you can't generally tell how much is lost as sound and how much as heat.

But with forces which are non - dissipative in nature (conservative forces) you don't have to worry about such losses and you have to just account for the Potential energy and kinetic energy which are easy to calculate.

So the thing is to recognise the force in action and then decide whether you should use it or not.

Improve this answer
$\endgroup$
0
2
$\begingroup$

You can use conservation of energy in any scenario where you can calculate the value of energy in all forms both at the beginning and the end of the time period in question.

You cannot use conservation of energy in problems where an unknown amount of energy is dissipated into the environment as heat, sound, or deformation of objects. This will be the case in any problem that involves friction or an inelastic collision. Note that energy is still conserved in these scenarios, but there is a term in the “before=after” energy equation that has an unknown value.

Improve this answer
$\endgroup$
0
1
$\begingroup$

Where to use conservation of energy?

You use conservation of energy whenever its applicable and its the most convenient method of calculating whatever you want to calculate. Conservation of energy is just another tool in your tool chest just like the equations of motion. Maxwell's' equation, Kirchhoff's' equation etc.

For most problems there are multiple ways to solve them. You can go with whatever feels easiest to you. This will depend on the problem, your skill sets and your preferences.

"Conservation of energy" is in many problems a good one to start with since the math tends to be on the easier side. If in doubt, try a few methods and choose the one that feels easiest to you.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I beliebe energy conservatipn is an extended application of equation of kinematics v2 - u2 = 2as,with only one form of energy that is 1/2 mv2 and that is only called kinetic energy and energy as such is not a broad and new thing,it just talks about v or distance,so any unknown in v2 - u2 = 2as should enable to use energy conservation. $\endgroup$
    – user201545
    Commented Aug 24, 2022 at 12:38
0
$\begingroup$

In the universe, “Conservation of total energy” holds.

If “the system is appropriately chosen (so that there is no addition or removal of matter and energy) and that you are able to account for all but one of parameters that appear in the resulting equation, thus leaving one parameter to solve for”,
then “Conservation of total energy” is useful…and we can use it to help solve our problem. (At this point, one should consider various example situations.)

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ that means the same thing that if time is asked it cant be solved by energy conservations,moreover i dnt know how to choose a syatem. $\endgroup$
    – user201545
    Commented Aug 24, 2022 at 5:29
  • $\begingroup$ @sachin It’s not the same thing. Some problems involving time can be solved with conservation of energy, and some can’t. The key is whether you have information to account for all the energies in your chosen system. Choosing the system is important (and is a follow up of your initial question). $\endgroup$
    – robphy
    Commented Aug 24, 2022 at 12:47
0
$\begingroup$

For any closed system where no energy is being added or taken out, the conversation of energy holds. In most high school/low undergrad physics problems, the systems studied are mechanical systems like blocks, particles, etc. under the influence of forces including mechanical forces, gravity, electromagnetism, etc.

Energy is conserved under the influence of gravitation and electromagnetism because they are conservative forces, which is something you may want to look into.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ if a block is sliding on an inclined plane,we take both the block and incline as a system,why dnt we take the ground or anything else into the system. $\endgroup$
    – user201545
    Commented Aug 24, 2022 at 9:48
  • $\begingroup$ you can, of course, but physics is a tool we have to understand and predict results to a problem, how would it help you to consider the ground into the problem, other than needlessly complicated it? $\endgroup$
    – Hritik Narayan
    Commented Aug 24, 2022 at 10:19
  • 1
    $\begingroup$ its not complicating but understanding,i believe the ground does not move,so not taken into. $\endgroup$
    – user201545
    Commented Aug 24, 2022 at 10:40