Work-energy theorem and Newton's second law

Question

I was going through a book in which it was written that

'Work-energy theorem for variable force, in general, does not incorporate the complete dynamics of Newton's 2nd law of motion as the theorem involves an integral over an interval of time. So temporal information contained in statement of 2nd law is integrated over and is not available explicitly.'

I am am understanding it to some extent but not fully. Can someone please simplify it or explain it the other way?

Answer 1

The work-energy theorem states that the net work done on a body is equal to it's change in kinetic energy: $W_\text{net}=\Delta K=K_\text{final}-K_\text{initial}$. This means it only deals with two instants in time: whatever you label as "initial" and whatever you label as "final". It tells you nothing about what happens at times between these two instants; the dynamics are lost, and the equation doesn't explicitly concern itself with the time over which the work is done. To determine "the rest" you would need to determine the dynamics using Newton's second law.

Answer 2

There are a lot of different derivations of the Work-Energy theorem out there, so to avoid ambiguity I present a derivation of the Work-Energy theorem that I will use as basis.

With that established I will address your question specifically.

I will use:
t time
s position
v velocity
a acceleration

$$ v = \frac{ds}{dt} \quad \Leftrightarrow \quad ds = v \ dt \tag{1} $$

$$ a = \frac{dv}{dt} \quad \Leftrightarrow \quad dv = a \ dt \tag{2} $$

The starting point is Newton's second law:

$$ F = ma \tag{3} $$

The next step is to integrate with respect to the spatial coordinate, integrating from starting point $s_0$ to final point $s$

The idea here is to carry the physics content of $F=ma$ forward by integrating both sides of (3).

$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{4} $$

We proceed to work out the right hand side. I omit the factor $m$ temporarily, it is a multiplicative factor that is just carried over each step

$$ \int_{s_0}^s a \ ds \tag{5} $$

Use (1) to change the differential from $ds$ to $dt$. Since the differential is changed the limits change accordingly.

$$ \int_{t_0}^t a \ v \ dt \tag{6} $$

Change the order:

$$ \int_{t_0}^t v \ a \ dt \tag{7} $$

Change of differential according to (2), with corresponding change of limits.

$$ \int_{v_0}^v v \ dv \tag{8} $$

So we have:

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{9} $$

We multiply both sides with $m$, and then the right hand side of (9) gives us the right hand side of (4). The result: the Work-Energy theorem:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{10} $$

So:
On the left hand side we have the expression for work done, and on the right hand side the expression for kinetic energy.

(As we know: the integral $\int_{s_0}^s F \ ds $ is well defined only if the outcome of the integration is independent of how the test object moves from start point to end point. A force with that property is referred to as a conservative force.)

As we know: potential energy is defined as minus work done.

Restating (10) in terms of energy:

$$ -\Delta E_p = \Delta E_k \tag{11} $$

Now: I have used the symbol $\Delta$ here, and the convention is that we use $\Delta$ for an interval with a definite size.

Here is where I start addressing your question specifically:
Quoting from that NCERT physics class 11 book:

[...] the theorem involves an integral over an interval [...]

Here is an important thing, that is not appreciated by the author of that book:

The validity of (11) extends all the way to infinitismally small increments:

$$ -d E_p = d E_k \tag{12} $$

Sure: (10) expresses an integral over a distance from a start point $s_0$ to a end point $s$. However, you can bring $s_0$ and $s$ arbitrarily close together.

So there is no mathematical obstacle to giving the following differential equation:

$$ - \frac{d E_p}{ds} = \frac{d E_k}{ds} \tag{13} $$

(13) says the following:
As a test object is being accelerated by a force the rate at which the kinetic energy is changing will match the rate at which the potential energy is changing.

That is:
After using the Work-Energy theorem to convert to potential and kinetic energy you can still evaluate the motion in terms of differentials.

The evaluation in terms of differentials will allow you to trace the motion from instant to instant.

(Remark: (13) is stated as differentiation with respect to the position coordinate $s$. Differentiation with repect to time is a valid equation too, but here differentiation with respect to the spatial coordinate is the efficient thing to do. Potential energy is the integral of force with respect to the spatial coordinate. So: differentiating potential energy with respect to the position coordinate recovers the force.)

Answer 3

When force varies, you need to integrate force over displacement to get the total work done: $$ W=\int _{a}^{b}\mathbf {F(s)} \cdot d\mathbf {s}\tag 1 $$

For example. Let's say we need to find a work done, when we compress a spring from $0$ to a final position $\ell$.

Spring force needed to compress to $x$, by Hooke's law, is $$ F(x)=kx \tag 2$$

Now substitute (2) into (1) : $$ \begin{align} W &= \int_0^\ell F(x)dx \\ &= \int_0^\ell kx~dx \\ &=\frac {k~\ell^{~2}}{2} \end{align} \tag 3 $$

After integration, notion of a force at a specific point $F(x)$ in spring, is lost. We are left only with total displacement $\ell$, hence final formula of total work done doesn't care about specific application of force at particular $x \in [0,\ell]$ path point (or at particular time moment). This temporal/spatial information is lost.

You can say to some sense, that varied force is averaged over the entire path by integration procedure. Actually, if you'll equate (3) to a work done by an average force:

$$ \frac {k~\ell^{~2}}{2} = \overline F ~\ell \tag 4,$$

you can extrapolate what is this average force in spring compression $ 0 \to \ell$ :

$$ \overline F = k~ \frac {\ell}{2} \tag 5 .$$

So, for getting total work done, you can either:

• integrate varied force over displacement or time, if over time then you need in eq (1) to change integration variables by : $W=\int _{a}^{b} {F(s)} \cdot ds = \int _{t_1}^{t_2} {F(t)} \cdot \frac {ds}{dt} dt = \int _{t_1}^{t_2} {F(t)} \cdot v(t) dt$

• or you just calculate average/effective force $\overline F$ over path and use it as in normal work done scenario.

Anyway, after integration or using an average force, temporal information like $F(t)=ma(t)$ is lost.

Answer 4

It is very simple so please do not complicate things here. The work-energy theorem is an integral equation $$\int^{T_f}_{T_i}\vec{F}\cdot d\vec{s}=K_f-K_i$$ which tells you what happens to the change in the object's kinetic energy after the interval $t=T_i$ to $t=T_f$. In particular one could measure the initial kinetic energy and the final kinetic energy and understand how much energy the force did work to supply to the object. Without knowing the specific force variable then, one has no idea what happens at $t\in(T_i,T_f)$. In general, integral theorems (principles) lose information at infinitesimal points in the interval integrated over, as we are more focused on the larger picture here.

Newton's second law as you see is a differential principle, it tells you at every instant of time what the rate change of motion of an object is based on the instantaneous force acting on the object.

Answer 5

Work-energy theorem, or whatever you call it, can't be equivalent to Newton's second law in general, since the former is a scalar equation, while the latter is a vector equation and thus contains more than one scalar equation in general.

This is not due to the fact that energy theorem is integrated in time, since it can be written in the differential form

$\dfrac{d K }{dt} = P^{tot}$,

being $K$ the kinetic energy of the system and $P^{tot}$ the total power of all the forces and monents acring on the system.

This is due to the different amount of quantity of information contained in a scalar (energy theorem) and in a vector equation (second law of dynamics): an implication exists and goes from the second law to the energy theorem (that is a theorem indeed, and not a fundamental principle, since it's one of their direct consequences), while it's not true an implication going in the other direction.

They could be equivalent only if the system has one degree of freedom.

Answer 6

Newton's second law $$F=m\dfrac{d^2x}{dt^2}\tag{1}$$

is a differential equation for the position of the particle $x(t)$. As such if you are able to solve it explicitly, you will be able to evaluate the position of the particle for each instant of time, $x(t)$, and from it compute many other observable quantities. The important point for your question is that you will be able to obtain all this information for all times.

On the other hand, the work-energy theorem relates two specific points in time. Indeed, recall that work is $$W=\int_{t_1}^{t_2} F dx.\tag{2}$$

The work-energy theorem then tells you that $$W=\dfrac{1}{2}mv_2^2-\dfrac{1}{2}mv_1^2\tag{3}.$$

So the particle is initially at $x(t_1)$ with velocity $v(t_1)$ and is found at $x(t_2)$ with velocity $v(t_2)$. What equation (3) allows you to do is to relate quantities at these two instants of time. So using it you will not be able to access observables at all times as solving Newton's second law for $x(t)$ would.

Answer 7

The main point to note is that when we do the general derivation for work energy theorem, we do a definite integral. When we do a definite integral of a function over a variable, the result is an expression independent of that variable. Consider for eg:

$$y(a,x)= ax$$

Where $a$ is a constant, if we integrate from $x=0$ to $x=1$, we have:

$$ G(a)= \int_{x=0}^1 y(a,x) dx = a \frac{x^2}{2}|_{x=0}^{x=1} = \frac{a}{2}$$

Similarly in the work energy theorem, we can simplfy the expression to result in an integral with time.

$$ \int F \cdot dx = \int \left( ma \cdot v \right) dt $$

So, clearly when we do the last integral over some bounds, we arrive at an expression which would have no "time" in it. Note that the simplification I did of:

$$ dx =v dt$$

Is not reliant on the thing I am integrating being $ma$, it could have been any expression for force.

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Example

Consider particle falling down,

$$ ma = -mg \hat{z}$$

we have:

$$ \int_{t=t_0}^{t=t_f} ma \cdot dx = \int_{t=t_0}^{t=t_f} -mg \hat{z} \cdot dx$$

$$ m \frac{\left(v(t_f) \right)^2}{2}- m \frac{\left(v(t_o) \right)^2}{2} = -mgh(t_f) - mgh(t_o)$$

You can see in the expression above that time is plugged in into the functions already. This is what the paragraph tells us.

Additionally, what we can do to make this more useful is to rearrange all the terms which are evaluated at $t_f$ to one side, we have:

$$ E= m \frac{\left(v(t_f) \right)^2}{2}+ mgh(t_f) = m \frac{\left(v(t_o) \right)^2}{2}+ mgh(t_o)$$

So, our conclusion in this case would be that the sum of potential and kinetic energy evaluated at a time $t$ is same for all time $t$.