Intuition for energy

Question

I was wondering how one might intuitively think of energy, especially in a 'compare and contrast' relationship with force. I tried to think of force as the rate of change of momentum, but was unable to think of a solid intuition for energy in particular. I'm just in high school so I will probably not understand more advanced concepts, is there any easy and intuitive way to understand 'energy'?

Answer 1

For a rough intuitive idea you can think of energy as "stored" motion. Or "stored" momentum if you will. "Stored" until "released".

• Energy is that which, when released, can cause momentum transfer, which
• creates forces, which
• causes acceleration, which
• is velocity changes.

This intuitive idea of energy as "stored" motion applies roughly to all types of energy, and the transfer of energy (moving the motion "storage" somewhere else") is via heat or work.

Although lacking detail and precision, this is the best attempt of an intuitive picture of energy that I have come across.

Answer 2

At the simplest level, energy is a mathematical construct with the nice properties of being conserved, which makes calculations easier. It is not to be compared with force directly, as energy is a combination of different factors, including force. Energy does tell you how much a force can change an object's motion, but that is only part of it and from a different perspective that the $F=ma$ relationship.

More in general, you can think of energy as a property of an object which can be stored in different forms with the constraint that the sum is constant. Suppose an object has three ways to store energy (the main ones in high school are: kinetic energy, gravitational energy, dissipation/friction/heat). For the sake of simplicity, let's call them $K$, $U$ and $F$. So you know that the total energy of the object is $E$ and that $E=K+U+F$. Now suppose $E=10$ in some units. Suppose you can measure $K=3$ and $U=5$ but you don't know $F$. Using the constraint that $K+U+F=10$ you can easily find $F=10-3-5=2$. So by knowing that energy is conserved, you can reduce the number of variables you need because the last one is specified by the conservation of energy.

Somebody in thé comments suggested to read Feynman's lectures: in there he makes a simple example of having $N=10$ blocks of something. If you one day check and only find 6 blocks, then (because you know that the total number is conserved) you know that 4 blocks are missing and you go look for them: you might not know where they are, but you know they are missing!

However, why is energy conserved?

The precise answer might be complicated, but at your level we can derive something from Newton's equation.

(Note: I will assume you don't know integrals, so I will do the calculations at finite interval and this will lead to some errors. If you know derivatives/integrals, you can try to do it again and see that you will get the classic result of the kinetic energy and of work. Otherwise please bear in mind that this is a very qualitative description! It's also a simplified 1D description. The concepts however remain valid!).

You know that $F=ma$ and now assume $F$ is constant. You can rewrite it as $F=m\Delta v / \Delta t$ and $\Delta v = \Delta x / \Delta t$ so we end up with

$$F=m \Delta x / \Delta t^2$$

We multiply everything by $\Delta x$ and get

$$F\Delta x = m (\Delta x / \Delta t)^2$$

Again, remeber that $$ \Delta x / \Delta t = \Delta v $$ so that we can write $$F\Delta x = m \Delta v^2$$ What do we mean by the symbol $\Delta$? Of course it is the difference between the value of the quantity at two different times $t_1$ and $t_2$. So that we can rewrite it as:

$$m \left(v^2(t=t_2)-v^2(t=t_1)\right) = F \left(x(t=t_2)-x(t=t_1)\right)$$

and rearrange it as

$$ m v^2(t=t_1)-F x(t=t_1) =m v^2(t=t_2)-F x(t=t_2)$$ so that the quantity $$E(t)=m v^2(t)-F x(t) $$ is constant over time, meaning that whatever time we choose, the sum of $m v^2$ and $-Fx$ is constant. We call this quantity energy! (Note again: because of small mistakes using finite quantities, this expression for energy is slightly wrong!)

Why is it useful? If we know now that an object of mass $m$, on which a force $F$ is acting, is moving at a speed of $1 m/s$ and we want to know how fast it will be moving after 10 seconds when it has reached a given position $x(t=10s)$, we can say: well, energy must be conserved so:

$$ m v^2(t=10s)-F x(t=10s) =m v^2(t=0s)-F x(t=s)$$

If we know the initial conditions, e.g. $x(t=0s)=0$ and $v(t=0)=1$ we get:

$$ m v^2(t=10s)-F x(t=10s) =m (1 m/s)^2$$ so that $$ v(t=10s)= \sqrt{m (1 m/s)^2+F x(t=10s) \over m}$$

If we know the force $F$, the mass $m$ and $x(t=10s)$ we can compute the position. This is also valid is $F$ is the sum of multiple forces. So using conservation of energy one can make faster calculations!

Or assume your objects is moving and at some point stops: then, by conservation of energy, we know some force was acting on it to stop it and we can use it to compute it. We just say $v(t=0)=v_0$, $x(t=0)=x_0$, $v(t=final \; time)=0$, $x(at\;which;the\;object\;stops)=x_f$ then $$E=m v^2(t=10s)-F x(t=10s) =m v^2(t=0s)-F x(t=s) = $$ $$= 0 - Fx_f=mv^2_0-Fx_0$$ so that $$F=-mv_0^2/(x_f-x_0)$$

Once again, using conservation of energy, we solved our problem. It's just a very useful quantity!

We could have derived similar results from the equation of motions, but it might have been much harder!

What did we learn? If we assume Newton's equation are right, we find that there exist a quantity, called energy, that is conserved over time, and we use it to make our computations easier. It's just a consequence of Newton's equations and it's useful!

However, as you will study further, you will find that the "conservation of energy" is a much more fundamental concept and that actually the equation of motions are a consequence of the conservation of energy- but it does not change that $energy$ is a quantity that things have and that they can use as a currency to do things, like moving, getting hot, etc.

It is a bit like mass: you have a definite amount of mass, you can do some things with (break something in two, hide some of it somewhere, convert it to another element) it but the total amount of mass is constant!

Answer 3

When I was in high school and I asked this question to my professor, he answered me something like "energy is just a quantity which has the nice property to be conserved", that's it. Maybe it is a little abstract, but I think it is a good way to think about it. And when you will go a little further in physics and learn about the Noether theorem and lagrangians, you will understand that this is in fact how we define energy : the conserved quantity stemming from time invariance of your theory.

Concerning its relationship with forces, have you already heard about the concept of work ? This is pretty what makes the link between force and energy. As stated on this Wikipedia article:

work is the energy transferred to or from an object via the application of force along a displacement

Answer 4

The most direct way is to see them as integral-derivative relationships.

Consider,

$$ K= \frac{1}{2} mv^2$$

Now, if you took the positive derivative:

$$ \frac{dK}{dx}= \frac{1}{2} m 2v \frac{dv}{dx}$$

Or,

$$ \frac{dK}{dx} = mv \frac{dv}{dx}$$

Now, consider the quantity:

$$ a(t)=\frac{ d v(x(t) )}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v$$

Hence,

$$ \frac{dK}{dx} = ma$$

It's well known that momentum's time derivative is force, but the procedure outlined above gives the answer to the question "What quantity's position derivative is force?". The good point about these energy relations is that you can ask questions such as "How do the forces change as I change the configuration of my system?" while if you simply stuck to newton's laws, you could only ask questions such as " How would my system evolve with time if I just let it go and do its own thing?"(*)

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To finish, I quote one of my favorite paragraphs from my high school physics textbook:

"While the WE theorem is useful in a variety of problems, it does not, in general, incorporate the complete dynamical information of Newton’s second law.

It is an integral form of Newton’s second law. Newton’s second law is a relation between acceleration and force at any instant of time. The work-energy theorem involves an integral over an interval of time. In this sense, the temporal (time) information contained in the statement of Newton’s second law is ‘integrated over’ and is not available explicitly.

Another observation is that Newton’s second law for two or three dimensions is in vector form whereas the work-energy the theorem is in scalar form. In the scalar form, information with respect to directions contained in Newton’s second law is not present"- NCERT class-11 physics, pg-119

However, I did not like the motivation used for introducing kinetic energy in that book. A better one is done by Ron Maimon in this post here

*: To work more with what happens as we change the configuration, it is useful to introduce potentials to begin.