I want direct answer for this and some interpretation with example. why do we need conservative force to define potential energy? what is wrong with non-conservative force and other? I have seen many question here but i don't understand.

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1Possible duplicates: https://physics.stackexchange.com/q/225594/2451 , https://physics.stackexchange.com/q/122345/2451 , https://physics.stackexchange.com/q/31672/2451 and links therein. – Qmechanic Oct 21 '17 at 19:02
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1Possible duplicate of What causes a force field to be "non-conservative?" – SchrodingersCat Oct 21 '17 at 21:28
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What is your proposed definition? – WillO Oct 22 '17 at 01:13
2 Answers
why do we need conservative force to define potential energy?
Conservative fields (and consequently conservative energies) are defined through a potential, known in this context as a scalar potential! A conservative field is a field whose integral is a potential, i.e.
$$\vec{F}=-\vec{\nabla} U$$
where $F$ is a field (e.g. electric field, gravitational field,...), $U$ is the potential, $\nabla=(\partial_{1},\dots,\partial_{N})$ is the vector of the N partial derivatives and the minus sign is put by convention.
From this definition you can develop the very useful property that a field is conservative iff the work employed for moving from two points in space does not depends on the path $\gamma$ chosen but it's only a function depending on the initial and final point.
In fact, wnce we have defined two points $A,B\in \mathbb{R}^3$ and a path $\gamma$ connecting them such that $$\gamma \colon [0,T]\to \mathbb{R}^3$$ $$t\mapsto \gamma(t)$$ and $\gamma(0)=A,\gamma(T)=B$, we have that $$\int_{\gamma}F\cdot d\mathbf{r}= \int_0^T F(\gamma(t)) dt= \int_0^T\nabla U(\gamma(t))dt= \big[U(\gamma(t))\big]_0^T=U(\gamma(B))-U(\gamma(A))=U(B)-U(A). $$
From this definition follows all the properties we are used to:
- Line integrals of $\textbf{F}$ are path independent.
- Line integrals of $\textbf{F}$ over closed loops are always 0.
- $\textbf{F}$ is the gradient of some scalar-valued function, i.e. $\textbf{F} = \nabla U$, for some function $U$.
- There is also another property equivalent to all these: $\textbf{F}$ is irrotational, meaning its curl is zero everywhere (with a slight caveat about the domani of definition).
what is wrong with non-conservative force and other?
Nothing wrong with them! The only problem is that all the properties listed above are not valid!
The gravitational force, spring force, magnetic force (according to some definitions) and electric force (at least in a time-independent magnetic field) are examples of conservative forces, while friction and air drag are classical examples of non-conservative forces.

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A non-conservative force does not "build up" or "store" energy which can later be released.
Friction is one. The work it does is not stored in any way. It is converted into heat and disappears.
Stored energy is what we call potential energy. Because it has a "potential" to do work, if released again.

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