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Mathematics for Physics

117 Newtonian Kinetic Energy Formula Derivation

A derivation of the Newtonian kinetic energy formula that is used to obtain \(K = \dfrac{1}{2}mv^{2}\) from the work-energy theorem.

Note: Also called \(T\) in some sources.

definition [d] (Newtonian Kinetic Energy Formula Derivation) From Shankar: start from the constant-acceleration kinematic relation

  • \(v_{2}^{2} = v_{1}^{2} + 2ad\) ,

with Newton’s second law \(a = \dfrac{F}{m}\), so

  • \(v_{2}^{2} = v_{1}^{2} + 2\dfrac{F}{m}d\) .

Rearranging gives

  • \(\dfrac{1}{2}mv_{2}^{2} - \dfrac{1}{2}mv_{1}^{2} = Fd\) .

Defining \(K = \dfrac{1}{2}mv^{2}\) and \(W = Fd\) yields the work-energy theorem

  • \(K_{2} - K_{1} = W\) .

where

  • \(v\) is the speed.
  • \(v_{1}\) and \(v_{2}\) are the initial and final speeds.
  • \(a\) is the acceleration.
  • \(d\) is the distance traveled.
  • \(F\) is a constant force.
  • \(m\) is the mass.
  • \(K\) is the kinetic energy.
  • \(K_{1}\) and \(K_{2}\) are the initial and final kinetic energies.
  • \(W\) is the work done by the force.

definition [d] (Newtonian Kinetic Energy Formula Derivation) From Logan: begin with the damped oscillator equation

  • \(m\dfrac{d^{2}x}{dt^{2}} + \gamma\dfrac{dx}{dt} + kx = 0\) .

Multiply by the velocity \(\dfrac{dx}{dt}\):

  • \(m\dfrac{dx}{dt}\dfrac{d^{2}x}{dt^{2}} + \gamma\left(\dfrac{dx}{dt}\right)^{2} + kx\dfrac{dx}{dt} = 0\) .

By the chain rule,

  • \(m\dfrac{dx}{dt}\dfrac{d^{2}x}{dt^{2}} = \dfrac{d}{dt}\left(\dfrac{1}{2}m\left(\dfrac{dx}{dt}\right)^{2}\right)\) ,
  • \(kx\dfrac{dx}{dt} = \dfrac{d}{dt}\left(\dfrac{1}{2}kx^{2}\right)\) ,

so

  • \(\dfrac{d}{dt}\left[\dfrac{1}{2}m\left(\dfrac{dx}{dt}\right)^{2} + \dfrac{1}{2}kx^{2}\right] = -\gamma\left(\dfrac{dx}{dt}\right)^{2}\) .

The kinetic energy term identified in the bracket is

  • \(T = \dfrac{1}{2}m\left(\dfrac{dx}{dt}\right)^{2}\) .

where

  • \(x\) is the displacement.
  • \(x(t)\) is the displacement as a function of time.
  • \(\dfrac{dx}{dt}\) is the velocity.
  • \(\dfrac{d^{2}x}{dt^{2}}\) is the acceleration.
  • \(m\) is the mass.
  • \(\gamma\) is the damping constant.
  • \(k\) is the spring constant.
  • \(t\) is time.
  • \(\dfrac{d}{dt}\) is the time derivative.
  • \(T\) is the kinetic energy.
  • \(\dfrac{1}{2}kx^{2}\) is the elastic potential energy in the oscillator.

117.1 Elementary Example

117.1.1 Simple

A constant force \(F = 6\,\mathrm{N}\) acts through \(d = 2\,\mathrm{m}\) on \(m = 3\,\mathrm{kg}\) starting from rest.

\[ W = Fd = 12\,\mathrm{J} \]

\[ K_{2} = W = \dfrac{1}{2}(3)v_{2}^{2} = 12\,\mathrm{J} \]

where

  • \(F\) is the constant force.
  • \(d\) is the distance traveled.
  • \(m\) is the mass.
  • \(W\) is the work done by \(F\).
  • \(K_{1}\) is the initial kinetic energy, equal to zero at rest.
  • \(K_{2}\) is the final kinetic energy.
  • \(v_{2}\) is the final speed.

117.1.2 General

For \(m = 2\,\mathrm{kg}\) and \(v = 4\,\mathrm{m/s}\),

\[ K = \dfrac{1}{2}mv^{2} = 16\,\mathrm{J} \]

where

  • \(m\) is the mass.
  • \(v\) is the speed.
  • \(K\) is the Newtonian kinetic energy.