How to Derive the Formula for Kinetic Energy

Begin with the Work-Energy Theorem., Rewrite work as an integral., Rewrite force in terms of velocity., Rewrite the integral in terms of a velocity differential., Integrate over change in velocity.

5 Steps 1 min read Medium

Step-by-Step Guide

  1. Step 1: Begin with the Work-Energy Theorem.

    The work that is done on an object is related to the change in its kinetic energy. ΔK=W{\displaystyle \Delta K=W}
  2. Step 2: Rewrite work as an integral.

    The end goal is to rewrite the integral in terms of a velocity differential. ΔK=∫F⋅dr{\displaystyle \Delta K=\int {\mathbf {F} }\cdot {\mathrm {d} }{\mathbf {r} }} , Note that mass is a scalar and can therefore be factored out. ΔK=∫ma⋅dr=m∫dvdt⋅dr{\displaystyle {\begin{aligned}\Delta K&=\int m{\mathbf {a} }\cdot {\mathrm {d} }{\mathbf {r} }\\&=m\int {\frac {{\mathrm {d} }{\mathbf {v} }}{{\mathrm {d} }t}}\cdot {\mathrm {d} }{\mathbf {r} }\end{aligned}}} , Here, it is trivial, because dot products commute.

    Recall the definition of velocity as well. ΔK=m∫drdt⋅dv=m∫v⋅dv{\displaystyle {\begin{aligned}\Delta K&=m\int {\frac {{\mathrm {d} }{\mathbf {r} }}{{\mathrm {d} }t}}\cdot {\mathrm {d} }{\mathbf {v} }\\&=m\int {\mathbf {v} }\cdot {\mathrm {d} }{\mathbf {v} }\end{aligned}}} , Typically, initial velocity v0{\displaystyle v_{0}} is set to
    0. ΔK=12mv2−12mv02=12mv2{\displaystyle {\begin{aligned}\Delta K&={\frac {1}{2}}mv^{2}-{\frac {1}{2}}mv_{0}^{2}\\&={\frac {1}{2}}mv^{2}\end{aligned}}}
  3. Step 3: Rewrite force in terms of velocity.

  4. Step 4: Rewrite the integral in terms of a velocity differential.

  5. Step 5: Integrate over change in velocity.

Detailed Guide

The work that is done on an object is related to the change in its kinetic energy. ΔK=W{\displaystyle \Delta K=W}

The end goal is to rewrite the integral in terms of a velocity differential. ΔK=∫F⋅dr{\displaystyle \Delta K=\int {\mathbf {F} }\cdot {\mathrm {d} }{\mathbf {r} }} , Note that mass is a scalar and can therefore be factored out. ΔK=∫ma⋅dr=m∫dvdt⋅dr{\displaystyle {\begin{aligned}\Delta K&=\int m{\mathbf {a} }\cdot {\mathrm {d} }{\mathbf {r} }\\&=m\int {\frac {{\mathrm {d} }{\mathbf {v} }}{{\mathrm {d} }t}}\cdot {\mathrm {d} }{\mathbf {r} }\end{aligned}}} , Here, it is trivial, because dot products commute.

Recall the definition of velocity as well. ΔK=m∫drdt⋅dv=m∫v⋅dv{\displaystyle {\begin{aligned}\Delta K&=m\int {\frac {{\mathrm {d} }{\mathbf {r} }}{{\mathrm {d} }t}}\cdot {\mathrm {d} }{\mathbf {v} }\\&=m\int {\mathbf {v} }\cdot {\mathrm {d} }{\mathbf {v} }\end{aligned}}} , Typically, initial velocity v0{\displaystyle v_{0}} is set to
0. ΔK=12mv2−12mv02=12mv2{\displaystyle {\begin{aligned}\Delta K&={\frac {1}{2}}mv^{2}-{\frac {1}{2}}mv_{0}^{2}\\&={\frac {1}{2}}mv^{2}\end{aligned}}}

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Rachel Richardson

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