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A few days ago I started studying classical mechanics and today I'm stuck with the integral formula for work. I didn't take the formula for granted and i tried to understand it, however, the presence of both dt and ds didn't allow me to grasp it deeply

$W=\int m\frac{d\mathbf{v}}{dt}\cdot d\mathbf{s}$

So I tried to elaborate this formula, and I get:

$dW=mvdv$

This means that the infinitesimal work is simply the dv of a body with momentum mv, and is very intuitive. Why in books the work formula is always presented as $W=\int \mathbf {F}\cdot d\mathbf{s}$ ? Wouldn't be better to explain first of all the momentum and then the fact that an infinitesimal change of energy is $dW=mvdv$ ?

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  • $\begingroup$ The basic definition of work is a force acting causes a displacement in its direction. This is from where the formula comes. It can be changed to other forms. $\endgroup$ – Tojrah Apr 26 at 15:53
  • $\begingroup$ the work done theorem is just a mathematical consequence of the newton's laws. but it has some deep implication. $\endgroup$ – Shing Apr 26 at 16:53
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What the other answers have failed to realize is that when you make the substitution $F=m\frac{\text d v}{\text d t}$ you are then talking about the net force, and hence the net work. However, often we want to look at the work done by a single force, and this requires you to use the definition of work done by a force

$$W=\int\mathbf F\cdot\text d\mathbf x$$

So it's not like one is better than the other. If you want to look at the work done by one force when multiple forces are acting on the object then you can't use $\text dW=mv\text dv$ to find the work done by that single force.

What you have given is just the work-energy energy theorem: $$\text dW=mv\text dv=\text d\left(\frac12mv^2\right)=\text dK$$ and this only deals with the net work done on the object. It cannot distinguish between the work done by various forces.

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Lets say I have a spring and I apply a force to slowly compress it. Now, nothing here has accelerated, so $mv\,\mathrm{d}v = 0$, but I certainly have done work on the spring. It is simply that I have changed the potential energy (and possibly thermal energy) rather than kinetic energy. This is why you need the expression in terms of the action of a general force.

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  • $\begingroup$ Why isn't the equation applicable in this scenario? $\endgroup$ – Aaron Stevens Apr 26 at 17:19
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Wouldn't be better to explain first of all the momentum and then the fact that an infinitesimal change of energy is $dW=mvdv$

To say the change in work with respect to velocity equals momentum is just another way to show the relationship between work and momentum. If you are more comfortable with starting from that vantage point, that’s OK. As @Tojah the expression for work can take many forms. Typically, the connection between work and momentum is as follows:

The change in momentum is

$$m\frac{dv}{dt}$$

Acceleration $a$ is

$$a=\frac{dv}{dt}$$

Force is $ma$ or

$$F=m\frac{dv}{dt}$$

Then work, which is force times distance in the direction of the force (dot product) then becomes

$$W=\int F.ds=\int m\frac{dv}{dt}.ds$$

Where $F$, $s$, and $a$ are vectors.

Hope this helps

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  • $\begingroup$ Let's only consider $dW=mvdv$, what is the physical meaning of $mvdv$? $\endgroup$ – Jhdoe Apr 28 at 17:48
  • $\begingroup$ @Jhdoe Why did you delete your last comment? Was working on a revision to answer it. $\endgroup$ – Bob D Apr 29 at 20:34
  • $\begingroup$ I didnt delete any comment .... $\endgroup$ – Jhdoe Apr 29 at 21:49
  • $\begingroup$ @Jhdoe Sorry, I thought you asked about conservative forces and their relationship to work. But, no problem. $\endgroup$ – Bob D Apr 29 at 22:20

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