Physics, asked by pallavi1039, 1 year ago

derivation of kinetic energy​

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Answered by Anonymous
4

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Answered by saanvigrover2007
13

 \mathfrak{Derivation \: of \:Kinetic \: Energy}

\mathsf{Things \: to \: know \: before\: Derivation}

 \mathsf{\implies Work done = Fs}

 \mathsf{\implies v² = u² + 2as}

 \mathsf{\implies s = \frac{v² - u²}{2a}}

 \mathsf{\implies u = 0 m/s \: for \:a \: body \: starting \: from \: rest}

 \mathsf{\implies Work \: Done = \: Energy}

 \mathsf{\implies Kinetic \: Energy \: is \: also \: written \: as \: E_k}

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

\mathsf{Derivation}

 \mathsf{\hookrightarrow E_k = Work done = Fs }

 \mathsf{\hookrightarrow \: = \: Fs \: = ma × s }

 \mathsf{\hookrightarrow E_k = m × \frac{v² - u²}{2a} × s}

 \mathsf{\hookrightarrow E_k = \frac{1}{2}mv²}

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