Question

please explain GRAVITATIONAL POTENTIAL ENERGY? AND PROVE THAT IT DOESN'T DEPENDS UPON PATH ?

Answer 1

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We all know instinctively that a heavy weight raised above someone's head represents a potentially dangerous situation. The weight may be well secured, so it is not necessarily dangerous. Our concern is that whatever is providing the force to secure the weight against gravity might fail. To use correct physics terminology, we are concerned about the gravitational potential energy of the weight.
All conservative forces have potential energy associated with them. The force of gravity is no exception. Gravitational potential energy is usually given the symbol U_gU​g​​U, start subscript, g, end subscript. It represents the potential an object has to do work as a result of being located at a particular position in a gravitational field.
Consider an object of mass mmm being lifted through a height hhh against the force of gravity as shown below. The object is lifted vertically by a pulley and rope, so the force due to lifting the box and the force due to gravity, F_gF​g​​F, start subscript, g, end subscript, are parallel. If gggis the magnitude of the gravitational acceleration, we can find the work done by the force on the weight by multiplying the magnitude of the force of gravity, F_gF​g​​F, start subscript, g, end subscript, times the vertical distance, hhh, it has moved through. This assumes the gravitational acceleration is constant over the height hhh.
$$\begin{array}{cc}{U}_g& =F_g\cdot h\\ & =m\cdot g\cdot h\end{array}$$​U​g​​​​​​=F​g​​⋅h​=m⋅g⋅h​​