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 :
- kinetic energy is proportional to $v^2$ and hence we cannot calculate change in energy as $(1/2)m(\Delta v)^2$
- 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)
Best Answer
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:
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 $v_1$.
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.