Physics, asked by MatheProBoy, 11 months ago

potential energy derivation​

Answers

Answered by laxmi1783
1

Answer:

In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors. Common types of potential energy include the gravitational potential energy of an object that depends on its mass and its distance from the center of mass of another object, the elastic potential energy of an extended spring, and the electric potential energy of an electric charge in an electric field. The unit for energy in the International System of Units (SI) is the joule, which has the symbol J.

Explanation:

Derivation for the equation of Potenstial Energy: Let the work done on the object against gravity = W. Work done, W = force × displacement. Work done, W = mg × h. Work done, W = mgh. Since workdone on the object is equal to mgh, an energy equal to mgh units is gained by the object . This is the potential energy (Ep) of the object.

(Electric potential energy

Electric potential energy, or electrostatic potential energy, is a potential energy (measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system. An object may have electric potential energy by virtue of two key elements: its own electric charge and its relative position to other electrically charged objects.)

Answered by saanvigrover2007
3

tex]\mathsf{Things \: to \: know \: before\: Derivation} [/tex]

 \mathsf{\implies Work \: done = Fs = F × height = \: Fh}

 \mathsf{\implies Work  = Energy}

 \mathsf{\implies Force = mass \: × \: acceleration \: = ma}

 \sf{\implies Acceleration \:due \:to \:gravity =\: 'g'}

 \mathsf{\implies Potential \: Energy \: is \: also \: written \: as \: E_p}

\mathsf{Derivation}

 \mathsf{\hookrightarrow E_p = Work done = Fs = Fh }

 \mathsf{\hookrightarrow E_p \: = \: Fh \: = ma × h }

 \mathsf{\hookrightarrow E_p\: = \: Fs \: = mg × h }

 \mathsf{\hookrightarrow E_p\: = \:mgh }

Similar questions