Rearrange the given steps to obtain the correct derivation of gravitational potential energy.
Answer variants:
Consider an object accelerating on a frictionless surface.
The gravitational force acting on the object is equal to its weight, \(F=mg\)
Hence, the expression for gravitational potential energy becomes \(PE\ =\ mgh\).
Substituting \(F\ =\ mg\) into the work done equation gives (W\ =\ mgh\)
Therefore, the work done is stored as kinetic energy.
Hence, the expression becomes, \(\frac{mg}{h}\)
Consider an object of mass m lifted vertically through a height \(h\).
Therefore, the work done is stored as gravitational potential energy.
The work done in lifting the object is given by \(W=F\times h\)
The correct order of derivation is:
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