How to Derive the Formula for Kinetic Energy

Reviewed by Joseph Meyer

Last Updated: November 23, 2016

When there are no opposing forces, a moving body tends to keep moving with a steady velocity as we know from Newton's first law of motion. If, however, a resultant force does act on a moving body in the direction of its motion, then it will accelerate per Newton's second law ${\mathbf {F}}=m{\mathbf {a}}.$ The work done by the force will become converted into increased kinetic energy in the body. We derive the expression for kinetic energy from these basic principles.

Method 1

Method 1 of 2:

Derivation Using Calculus

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1
Begin with the Work-Energy Theorem. The work that is done on an object is related to the change in its kinetic energy.
- $\Delta K=W$
2
Rewrite work as an integral. The end goal is to rewrite the integral in terms of a velocity differential.
- $\Delta K=\int {\mathbf {F}}\cdot {\mathrm {d}}{\mathbf {r}}$
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3
Rewrite force in terms of velocity. Note that mass is a scalar and can therefore be factored out.
- ${\begin{aligned}\Delta K&=\int m{\mathbf {a}}\cdot {\mathrm {d}}{\mathbf {r}}\\&=m\int {\frac {{\mathrm {d}}{\mathbf {v}}}{{\mathrm {d}}t}}\cdot {\mathrm {d}}{\mathbf {r}}\end{aligned}}$
4
Rewrite the integral in terms of a velocity differential. Here, it is trivial, because dot products commute. Recall the definition of velocity as well.
- ${\begin{aligned}\Delta K&=m\int {\frac {{\mathrm {d}}{\mathbf {r}}}{{\mathrm {d}}t}}\cdot {\mathrm {d}}{\mathbf {v}}\\&=m\int {\mathbf {v}}\cdot {\mathrm {d}}{\mathbf {v}}\end{aligned}}$
5
Integrate over change in velocity. Typically, initial velocity $v_{{0}}$ is set to 0.
- ${\begin{aligned}\Delta K&={\frac {1}{2}}mv^{{2}}-{\frac {1}{2}}mv_{{0}}^{{2}}\\&={\frac {1}{2}}mv^{{2}}\end{aligned}}$

Method 2

Method 2 of 2:

Derivation Using Algebra

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1
Begin with the Work-Energy Theorem. The work that is done on an object is related to the change in its kinetic energy.
- $\Delta K=W$
2
Rewrite work in terms of acceleration. Note that using algebra alone in this derivation restricts us to constant acceleration.
- ${\begin{aligned}\Delta K&=F\Delta x\\&=ma\Delta x\end{aligned}}$
- Here, $\Delta x$ is the displacement.
3
Relate velocity, acceleration, and displacement. There are several constant acceleration kinematic equations that relate time, displacement, velocity, and acceleration. The "timeless" equation which does not contain time is below.
- $v^{{2}}=v_{{0}}^{{2}}+2a\Delta x$
- When an object starts from rest, $v_{{0}}=0.$
4
Solve for acceleration. Remember, initial velocity is 0.
- $a={\frac {v^{{2}}}{2\Delta x}}$
5
Substitute acceleration into the original equation and simplify.
- ${\begin{aligned}\Delta K&=m\left({\frac {v^{{2}}}{2\Delta x}}\right)\Delta x\\&={\frac {1}{2}}mv^{{2}}\end{aligned}}$

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How do I calculate kinetic energy using the equations of motion?

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K=1/2mv^2. You can take any equation of motion, solve for v and then substitute in the equation above. I could provide more help if you gave the specific equation.

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How do I calculate potential and kinetic energy involving projectile of a ball without given mass?

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You can not, since energy is a function of both mass and velocity. If you know the velocity but not the mass, you can turn it into a function of mass, since velocity becomes a constant; you could have K = 7m, where m is the mass. Just be sure to mind your units.

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Why are we using the timeless equation?

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We're using the timeless equation because we know that the kinetic energy of something shouldn't depend on time. So by using this equation, we can arrive at the answer in the simplest possible way. Otherwise, we would have to find a way to cancel the time variable that would be introduced.

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