# Why is Energy gained /work done on object dependent on it's initial energy/momentum? Difference between Forces acting simultaneously and separately

Suppose an object starts from rest and attains a velocity $$v_1$$.Its kinetic energy (and also change in kinetic energy) is $$(1/2) mv_1^2$$.
Now suppose that the object starts with velocity $$v_0$$ and attains an extra velocity of $$v_1$$. Here, the change in kinetic energy is $$(1/2)m((v_0 + v_1 )^2 - v_0 ^2)$$ and not $$(1/2) mv_1^2$$.

However, momentum change in both the cases are equal. Why does the energy change depend on the initial energy/velocity/momentum it had? Mathematically, I understand that :

1. kinetic energy is proportional to $$v^2$$ and hence we cannot calculate change in energy as $$(1/2)m(\Delta v)^2$$
2. Work done is force times displacement and if the object starts at a higher speed, its displacement during the application of force will be higher, hence mathematics gives us a higher 'work done'

But I want to understand the physical significance. Why does the same momentum change not mean same energy change? Why does work done depend on the displacement of the object that was not caused entirely by the force but was due to the velocity it initially had?
(edit: extra displacement of $$v_0\Delta t$$, due not to the force F but due to the initial velocity $$v_0$$ it already had , in other words, the force that is supposed to be doing work is not the cause of the entire displacement, and yet we multiply it by the entire displacement 's'!! wouldn't it be more logical if the displacement was restricted to that caused by the force only? and this makes me question the definition of F.ds for work done!)

Another question that I would like to ask (a related question, or perhaps conceptually the same doubt just rephrased):

say a force of $$F_1$$ can cause a displacement of an object to the right. A force $$F_2$$ is capable of causing a leftward movement (consider constant forces for simplicity).
If I apply the forces separately, say the displacements are $$d_1$$ and $$d_2$$(consider the forces applied for the same time interval $$\Delta t$$).
If I apply them simultaneously, let the displacement be $$d_3 < d_1$$ and $$d_2$$

Work done by $$F_1$$ in case 1 is $$F_1.d_1$$
Work done by $$F_1$$ in case 2 is $$F_1.d_3$$ < $$F_1.d_1$$
Net work done on the object = $$(F_1 - F_2).d_3$$ and not $$F_1.d_1$$ - $$F_2.d_2$$

I am not being able to digest this, and feel like there is a loss of something somewhere... why is the work done by $$F_1$$ lesser in the second case just because of the existence of another force? shouldn't it be independent ? where did that difference in work done/energy go to?

To be more explicit, suppose $$F_1$$ acts for a time interval of $$\Delta t$$ and then after its work $$F_2$$ acts for the same time interval $$\Delta t$$. The momentum change would be $$F_1 \Delta t -F_2 \Delta t = (F_1-F_2) \Delta t$$ - same as what would happen if both forces acted simultaneously during the interval $$\Delta t$$. Yet, the work done (by each of the forces, and even the net work on the object) and hence the change in energy will be different.

Why so? What is happening to that difference in energy in the two cases... -same forces, same change in momentum, just a small difference in the experiment in terms of when the forces are applied...

(I would like to get an explanation of the physical interpretation, because the math has already been worked out and interpreting it is the problem)

• In common man's language, I can say that I observe that an opposing force is more effective in decreasing the energy of an object if it prevents the force against which it acts from increasing the objects energy rather than allowing the increase and then trying to decrease it! Commented Mar 25, 2022 at 13:41

TL;DR It seems to me that you are trying to find something more fundamental to derive definitions for work and momentum. There is nothing more fundamental than the second Newton's law of motion which directly relates force and momentum (or quantity of motion as Newton calls it)

$$\vec{F} = \frac{d \vec{p}}{dt} = \frac{d}{dt} m \vec{v}$$

Isaac Newton deduced the above equation (law) from a number of observations (experimental results) available to him at the time. It is proved to be true by many scientists after Newton doing even more experiments. If you are interested in discussions like this, then I would definitely recommend getting your hands on "The Feynman Lectures on Physics". The book (in 3 volumes) gives context for many principles that we often take for granted, such as what energy actually means. In short, it is just an abstraction proved to be useful! (See 4-1 What is energy? in Vol. 1)

Here is an interesting article on history of definitions for momentum (vis mortua, dead force) and kinetic energy (vis viva, living force): "D'Alembert and the Vis Viva Controversy" by C. Iltis.

The excerpt regarding kinetic energy:

Boscovich suggested that, if the time coordinate is replaced by the space traversed and the pressure coordinate by the force which at any instant produces the velocity proportional to it, a second aspect of the phenomenon is represented. Boscovich, however, explained neither this substitution nor the introduction of the concept of force. The new term 'force' must be interpreted as an entity proportional to the velocity engendered at any instant. If the pressure coordinate is changed to the force and the time coordinate to the space then the new geometrical image producing the velocity would be represented in modern notation as $$\int F ds$$. We would then interpret vis viva as $$\int m v dv = \int F ds$$ (where $$ds = v dt$$). Boscovich does not bring the mass into this analysis.

Why does the same momentum change not mean same energy change?

Simply because momentum and kinetic energy are not defined in the same way: (i) change of momentum is defined as force over time $$\Delta p = F \cdot \Delta t$$ (impulse-momentum theorem), while (ii) change of kinetic energy is defined as force over displacement (work-energy theorem) $$\Delta K = F \cdot \Delta s$$.

In your example, for the same acceleration $$a$$ (i.e. force $$F$$) it takes equal amount of time to accelerate from $$0$$ to $$v_1$$ or from $$v_0$$ to $$v_0+v_1$$

\begin{aligned} v_1 &= 0 + a \Delta t \\ v_1 + v_0 &= v_0 + a \Delta t \end{aligned}

However, the displacement in the latter case is greater by $$v_0 \Delta t$$

\begin{aligned} \Delta s &= \frac{1}{2} a (\Delta t)^2 + 0 \Delta t \\ \Delta s &= \frac{1}{2} a (\Delta t)^2 + v_0 \Delta t \end{aligned}

For the same force $$F$$ (i.e. acceleration $$a$$), $$\Delta t$$ is the same in two cases which means that change of momentum will also be the same, but the displacement $$\Delta s$$ is greater by $$v_0 \Delta t$$ in the second case which means that it takes more work to reach speed $$v_1$$.

Why so? What is happening to that difference in energy in the two cases... -same forces, same change in momentum, just a small difference in the experiment in terms of when the forces are applied...

In your second question you are trying to integrate the force over time, which is the definition for the impulse $$J$$ and not for the work $$W$$. If you still want to get the work by doing integration over time then you have to do the following:

$$W = \int F \cdot ds = \int F \frac{ds}{dt} dt = \int F v \cdot dt$$

In other words, you should actually integrate $$F v$$ over time instead of $$F$$. This is why it matters what is the object velocity when you apply force $$F_1$$ and $$F_2$$ in your second example.

• But it is because Work is defined in that way. But now this definition of work makes no sense to me because the extra displacement $v_0\Delta t$, as I had mentioned in the question is due to the already existing velocity and not due to the force exerted later, the force did not cause it. So I guess I end up questioning the very validity (and meaning and interpretation) of the definition of work. Commented Mar 23, 2022 at 16:53
• How was work derived to be F.ds? If it was defined to be so, it has to be done in a way that equals the change in energy, and perhaps many other factors for it to be meaning ful. Commented Mar 23, 2022 at 16:54
• @EternalLearner What is the true meaning of energy? Before you answer that question - what is the true meaning of force? These are just abstract concepts around which we are building our models of the real-world. I do not know of anything more fundamental to derive $W = \int F ds$ from. Note that the work-energy theorem $\Delta K = W$ is derived directly from the second Newton's law of motion. I would recommend reading the "The Feynman Lectures on Physics" Vol. 1 - it has discussions on the line what you are asking. Commented Mar 23, 2022 at 17:06
• Sure, thank you. I will get my hands on the book. I think your answer has guided me to the right path to get the complete answer . Commented Mar 25, 2022 at 13:07
• @EternalLearner I am glad it helped. Thanks for accepting the answer ;-) Commented Mar 25, 2022 at 17:44

Net force, as measured by a load cell, happens to be proportional to the derivative of momentum: $$\mathbf F = \frac{\mathbf{dp}}{dt}$$ It is a experimental result.

Making dot product at both sides with an infinitesimal displacement $$\mathbf{dr}$$:

$$\mathbf {F.dr} = \frac{\mathbf{dp}}{dt}\mathbf {.dr} = m\frac{\mathbf{dv}}{dt}\mathbf {.dr} = m\mathbf{dv.}\frac{\mathbf{dr}}{dt} = m\mathbf{v.dv} = md\left(\frac{1}{2}\mathbf{v.v}\right)$$

That implies: $$dw = \mathbf {F.dr} = d\left(\frac{1}{2}m|v|^2\right)$$

So, "work" and "kinetic energy" are names to mathematical defined quantities. While we are dealing with pure motion of objects there is nothing more than that.

The intuition behind is the extent of damage that an object can make in a collision. It is (experimentaly) proportional to the kinetic energy, and not to the momentum. And when we talk of a collision, that means with something at rest in the frame where the velocities are measured.

That way, the same force that changes the momentum by a certain amount can increase the damaging capability differently, depending on the initial velocity of the object.

• You mean ' rate of change of momentum' in your first line right? An edit could perhaps prevent other readers from getting confused... Commented Mar 25, 2022 at 12:47
• The information about the experimental observations is helpful, thank you. Commented Mar 25, 2022 at 13:29

What is happening to that difference in energy in the two cases... -same forces, same change in momentum, just a small difference in the experiment in terms of when the forces are applied...

Try a practical example. Go find one of those push merry-go-rounds on a playground.

Think of pushing on it with a 10N force. From rest, that will be easy. If I ask you to apply that force constantly for a full one second, that will also be easy.

Now, get the merry-go-round going quite quickly. If I ask you to "simply" apply that same 10N force, you will find it difficult or impossible. The energy required to do so is much greater. Further, even if you can manage to do so, you won't be able to sustain it for the full second.

At high relative velocity, it simply becomes more difficult to transfer the energy.

• I can agree to this example because, to apply force I will have to run behind the object that's already moving. That's because, it is a contact force and I, a living being will obviously find it difficult to run faster. The whole thing is because it is a contact force. However, what if I consider a charged point object in a uniform electric field? Commented Mar 25, 2022 at 11:57
• Electric fields don't appear spontaneously as a frame-independent existence. Just as the energy from light changes with redshift, the total force on the particle changes as the relative velocity changes. Commented Mar 25, 2022 at 15:53

Its really short and imaginary answer but i think it could help. Imagine you are accelerating through space-time. Your velocity and kinetic energy are increasing. Because of the energy and mass you are making curves in space-time. Unlike your momentum the curves of space-time are increasing exponentially . So it means you need more energy to get the same force.