Answer to Question #166471 in Mechanics | Relativity for Wajahat

Question

An object fall from rest at a height h Determine the potential energy of an object as a function of t

Answer 1

The potential energy of an object cannot be defined as a function of time.

We can prove this as follows:

By law of conservation of energy :

$E_{total}=T+U$

Case I :

Assuming potential energy $U$ is independent of time,

If the potential energy is independent of time, then the mechanical energy (T+U) is a constant of the motion; i.e., energy is conserved

Differentiating above equation with resect to time, t

$\Rightarrow\dfrac{dE_{total}}{dt}=\dfrac{dT}{dt}+\dfrac{dU\{r\}}{dt}$

$\Rightarrow \dfrac{dE_{total}}{dt}=\dfrac{1}{\cancel2}\cancel2mv\dfrac{ dv}{dt}+\dfrac{\partial U}{\partial r}.\dfrac{\partial r}{\partial t}$

$\Rightarrow\dfrac{dE_{total}}{dt}={v.F}+\nabla U.v$

$\Rightarrow\dfrac{dE_{total}}{dt}=\cancel{v.F}-\cancel{v.F}$

$\Rightarrow\dfrac{dE_{total}}{dt}=0$

$\therefore$ E is constant

Case II :

Assuming potential energy $U\{r,t\}$ is a function of time and position of the object

$\Rightarrow E_{total}=\dfrac{1}{2}mv^2+U\{r,t\}$

Differentiating above equation with respect to time, t

$\Rightarrow \dfrac{dE_{total}}{dt}=\dfrac{1}{\cancel2}\cancel2mv\dfrac{ dv}{dt}+\dfrac{\partial U}{\partial t}+\nabla U\dfrac{\partial r}{\partial t}$

$\Rightarrow\dfrac{dE_{total}}{dt}=\cancel{v.F}+\dfrac{\partial U}{\partial t}-\cancel{v.F}$

$\Rightarrow\dfrac{dE_{total}}{dt}=\dfrac{\partial U}{\partial t}$ which is not equal to 0

If U depends on time, then energy must be changing in other parts of the full system.

But since there is no external work done on the particle therefore total energy should be conserved and change with respect to time should be zero.