# Question regarding Conservative Forces

1. Work-Energy Theorem States that

$$dT=F.dr$$.

On integration we can see that

$$T_2 - T_1$$ = $$\int_C F.dr$$ (C is the curve along which the object moves)

Does this mean the force can be expressed as $$\nabla T$$ (T being the Kinetic Energy ) ?

If not why can't we write it that way ?

2)Work Energy Theorem is derived by differentiating Kinetic Energy with respect to time

$$\dfrac{dT}{dt}$$ = $$ma.v$$.

$$dT=F.dr$$

Now this change in Kinetic Energy is the change that the particle experience as time passes (this change is brought on by time dependency ) .

Now how can this result be used for conservative forces?

$$ΔΕ=ΔT+ΔU$$

T is the Kinetic Energy ,U is the potential energy .

Now $$ΔE= 0$$ (E isn't dependent on time for conservative forces )

But isn't change in Kinetic energy brought on by time? How can the result $$dT=F.dr$$ (a relation derived through time dependency ) be used in a time independent result .

• > 1) Work Energy Theorem States that > > $dT=F.dr$. Be careful. The theorem deals with net work. So you should write $$dT=F_{net}dr$$ Commented Apr 4, 2022 at 13:36

On integration we can see that

$$T_2 - T_1$$ = $$\int_C F.dr$$ (C is the curve along which the object moves)

Does this mean the force can be expressed as $$\nabla T$$ (T being the Kinetic Energy ) ?

Not necessarily. It is only the net force, and therefore the net work done, that can be expressed as $$\nabla T$$. That means you need to take into account all of the forces acting on the object moving along the curve and the work done by each. An that's because the work done by a particular force may be negative or positive.

For example, If I lift an object of mass $$m$$ from rest and bring it to rest at height $$h$$ there are two forces doing work on the object. There is mine which acts is in the same direction as the displacement of the object and which therefore does positive work of $$mgh$$. And there is gravity which acts opposite to the direction of the displacement therefore doing negative work of $$-mgh$$. The net work is zero and the change in kinetic energy is zero. When something does positive work on an object it transfers energy to the object. If it does negative work on an object it takes energy away from the object. In this example gravity takes the energy I gave the object and stores it as gravitational potential energy (GPE) of the object-earth system.

But isn't change in Kinetic energy brought on by time? How can the result $$dT=F.dr$$ (a relation derived through time dependency) be used in a time independent result .

I'm not quite sure what you are getting at, but in applying the work energy theorem you are only interested in the initial and final kinetic energy between two points. How the KE may vary in time in between is irrelevant.

Hope this helps.