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I have a doubt in this question, Diagram given below

Block of some mass falling down along a curvilinear path

In the Question it is asked that "calculate the final velocity of the block in the figure" and in the solutions it is given that the work done by normal force exerted by the surface is not considered when applying the work-energy theorem

but according to the proof of the work-energy theorem, the "'net' external force X displacement upto which the force was applied is equal to the change in kinetic energy of the object." (could use integration by taking the variable force to be constant for infinitely small time intervals)

but by this definition normal force should also be added to calculate the net external force on the object as it also always acts in opposite directions to gravity (with different magnitudes obviously)

So why is the normal force not considered here? Is the question/solution wrong?
If it is correct, why? And what is the mathematical proof of not considering the normal force when applying the work-energy theorem?

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  • $\begingroup$ “Do not close this question” is almost certainly going to get this question closed. $\endgroup$ Nov 11, 2023 at 12:33
  • $\begingroup$ i asked the same question few days ago, they closed it for some absurd reason $\endgroup$ Nov 11, 2023 at 12:38
  • $\begingroup$ Asking duplicate questions is not a good idea. $\endgroup$ Nov 11, 2023 at 12:40
  • $\begingroup$ I agree with @MattHanson that asking duplicate questions is inappropriate. But to me this question is not really a duplicate of the previous one. They only share the same drawing. I think it is ok $\endgroup$
    – Dale
    Nov 11, 2023 at 12:53

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You could indeed use the work-energy theorem to argue that the energy gained by the block is equal to the work done on it by the net external force acting on it, which is its weight $m\vec g$ plus the normal force $\vec N$. Note that both $\vec g$ and $\vec N$ are vectors and do not act in the same direction so you need to add them as vectors. Also note that the normal force acts perpendicularly to the surface, not in the opposite direction to gravity. So the net external force on the block is

$\vec F = m \vec g + \vec N$

The work done on the block when it moves a small distance $\vec {dx}$ is then

$dW = \vec F . \vec {dx} = (m \vec g + \vec N) . \vec {dx}$

Once again, note that $\vec {dx}$ is a vector, and to find the scalar quantity $dW$ then we must take the dot product or scalar product of $m \vec g + \vec N$ with $\vec {dx}$. Finally, we can integrate $dW$ across the whole motion to get

$\displaystyle W = \int dW = \int (m \vec g + \vec N) . \vec {dx}$

However, it is much simpler (and gives the same result) if we use conservation of energy to argue that the kinetic energy gained by the block is equal to the potential energy that it loses, which is $mgh$ where $h$ is the vertical distance that it has fallen.

To see that these two methods give the same result, notice that $\vec N$ is always perpendicular to $\vec {dx}$, so $\vec N . \vec {dx} = 0$ (in other words, the normal force does no work on the block) and we have

$\displaystyle W = \int (m \vec g + \vec N) . \vec {dx} = \int m \vec g . \vec {dx} + \int \vec N . \vec {dx} = m \int \vec g . \vec {dx} = mgh$

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  • $\begingroup$ oooohhh!!! Thank you so so so much for proving it using work-energy theorem to prove it, and also giving such a beautiful mathematical proof of it, thank you so so so much $\endgroup$ Nov 11, 2023 at 13:14
  • $\begingroup$ things are sometimes difficult to do when you always split everything into x and y components and then do the math. Thank you so much!! $\endgroup$ Nov 11, 2023 at 13:24
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Indeed, since the net force is not calculated this question does not use the work energy theorem. The work energy theorem speaks only of the “net work”, which is the work done by the net force. You cannot neglect a force that contributes to the net force in using it.

What should be used here is the conservation of energy. The gravitational potential energy decreases, so the kinetic energy increases. The normal force doesn’t have a change in potential energy, so it is neglected on that basis. Usually conservation of energy is more useful than the work energy theorem, in my opinion.

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  • $\begingroup$ but how can conservation of energy be used here as normal force( an external force) acting on it changes the total energy of the body ? $\endgroup$ Nov 11, 2023 at 12:36
  • $\begingroup$ @SauravMishra no, the normal force does not change the total energy. The ground does not have a potential or internal energy that changes in this problem. Where would it get the energy to change the total energy of the body? $\endgroup$
    – Dale
    Nov 11, 2023 at 12:40
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    $\begingroup$ oooohhhhh!!! yes you are right. in my country, conservation of energy is not present in class 11 syllabus in physics (for 16 year olds), we have just heard about it, so i had no idea about its proof and application. thank you so so so much for your help. $\endgroup$ Nov 11, 2023 at 12:52
  • $\begingroup$ @Dale I find it strange your justification for neglecting the normal forces. There are plenty of forces that don't have a potential/internal energy associated with them (i.e., non-conservative forces) that can change the energy of a system by doing work on the system. The reason we neglect the normal force here is not because "The normal force doesn’t have a change in potential energy, so it is neglected on that basis"; rather, it's because the normal force does no work because it is at all times perpendicular to the direction of motion. $\endgroup$
    – march
    Nov 11, 2023 at 21:32
  • $\begingroup$ @march I am not sure what you are objecting to. That is two ways of saying the same thing. When you move perpendicular to a force the potential energy doesn’t change $\endgroup$
    – Dale
    Nov 11, 2023 at 21:47
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In this answer I want to clarify the nature of the Work-Energy theorem. (The issue of the normal force exerted by the supporting surface has already been addressed in the answer by contributor Er Jio.)

To that end I will present a derivation of the Work-Energy theorem.

The starting point is Newton's second law:

$$ F = ma \tag{1} $$

The next step is to evaluate an integral: we integrate both sides of (1) with respect to the position coordinate, integrating from starting point $s_0$ to final point $s$

$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{2} $$

The work-energy theorem hinges on the fact that the two factors in the right hand side of (2), acceleration $a$ and position coordinate $s$, are not independent of each other; acceleration is the second derivative of position.

In preparation for developing the right hand side of (2) we need to do some mathematical operations.

In the steps starting with (5) the relations (3) and (4) will be used:

$$ v = \frac{ds}{dt} \ \Leftrightarrow \ ds = v \ dt \tag{3} $$

$$ a = \frac{dv}{dt} \ \Leftrightarrow \ dv = a \ dt \tag{4} $$

Going from (5) to (8):
First (3) is used to change the differential from $ds$ to $dt$, with corresponding change of limits. Next (4) is used - with change of limits - to arrive at (8).

$$ \int_{s_0}^s a \ ds \tag{5} $$ $$ \int_{t_0}^t a \ v \ dt \tag{6} $$ $$ \int_{t_0}^t v \ a \ dt \tag{7} $$ $$ \int_{v_0}^v v \ dv \tag{8} $$

The steps from (5) to (8) give the following mathematical relation:

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

Notice especially:
The definitions (3) and (4), in combined form, are all that it takes to imply (9). The reason that (9) obtains is purely mathematical: in any situation where you have a quantity $s$, its first derivative, and its second derivative: the relation expressed by (9) obtains.


We use (9) to go from (2) to the work-energy theorem:

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

We can think of the Work-Energy theorem as consisting of two components of different nature:

  • The mathematical property expressed by (9)
  • Newton's second law: $F=ma$

That is, the Work-Energy theorem consists of mathematical content and physics content, and the physics content is the relation $F=ma$



The left hand side of (10) is the definition of work done. As we know: change of potential energy is defined as the negative of work done.

$$ \Delta E_p = - \int_{s_0}^s F \ ds $$

The Work-Energy theorem, (10), is a statement in terms of integrals. We have that integration is inherently an evaluation from a starting point to an end point. That is, an integral gives a difference between a starting point and an end point. Therefore, if we shift the notation to $E_k$ and $E_p$ (kinetic energy and potential energy), we must do so in terms of a $\Delta$ of energy

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

It follows from (11) that as a system goes through dynamic change the amount of change of kinetic energy will always match the amount of change of potential energy.

$$ \Delta E_k + \Delta E_p = 0 \tag{12} $$

(12) is of course a statement of conservation; (12) implies conservation of the sum of kinetic energy and potential energy.

$$ E_k + E_p = C \tag{13} $$


We recognize the well known principle of conservation of Energy as a generalization of (12). In order for the principle of conservation of energy to hold good: all forms of potential energy must share this property.

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The normal force corresponds to the defining constraint of a "solid surface" that motion can only exist along the surface and can't be through the surface. According to this definition, the normal force always acts perpendicular to the surface to cancel out any external force that would accelerate a body into the surface. So the motion is always parallel to the surface, and the normal force is always perpendicular to the surface, then the velocity is always perpendicular to the normal force:

$$\mathbf F_N\cdot \mathbf v = 0$$

This expression is also equal to the instantaneous power transferred by the force, i.e. the rate of work done, so this means that the normal force does no work by definition. Therefore we can ignore it when considering the work-energy theorem since it contributes $0$ work anyways.

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  • $\begingroup$ oohhh!!! thank you so much for clarifying my other doubt i asked in the previous question, thank you so much $\endgroup$ Nov 11, 2023 at 13:28
  • $\begingroup$ but are you considering the normal force by the surface which only provides an constraint motion to the object? coz if you push 2 blocks in a series, the first one's movement is slowed by the normal force exerted by the second block $\endgroup$ Nov 11, 2023 at 13:38
  • $\begingroup$ @SauravMishra that's actually a good point which I hadn't considered. The consideration I made in my original answer was that the block is initially at rest in the direction perpendicular to the surface, and in that case the block always moves parallel to the surface and the normal force never does work. However if the block begins moving towards the surface, for example it is dropped at some height above the surface, then the normal force will in fact do work as the block bounces off. Again this is a very good point you brought up. $\endgroup$
    – Er Jio
    Nov 11, 2023 at 13:56

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