AcademyWork, Energy, and Power
Academy
Kinetic Energy and the Work-Energy Principle
Level 1 - Physics topic page in Work, Energy, and Power.
Principle
Net work changes translational kinetic energy.
Notation
\(K\)
translational kinetic energy
\(\mathrm{J}\)
\(m\)
mass
\(\mathrm{kg}\)
\(v\)
speed
\(\mathrm{m\,s^{-1}}\)
\(W_{\mathrm{net}}\)
total work by all forces
\(\mathrm{J}\)
\(\Delta K\)
change in kinetic energy
\(\mathrm{J}\)
Method
The work-energy relation comes from applying Newton's second law along the displacement and rewriting acceleration using speed.
Net work
\[W_{\mathrm{net}}=\int_{x_i}^{x_f}F_{\mathrm{net}}\,dx\]
Newton model
\[F_{\mathrm{net}}=ma\]
Substitute acceleration
\[a=\frac{dv}{dt}\]
Change variable
\[v=\frac{dx}{dt}\Rightarrow dx=v\,dt\]
Replace dx
\[a\,dx=\frac{dv}{dt}\,v\,dt=v\,dv\]
Integrate speed
\[W_{\mathrm{net}}=\int_{v_i}^{v_f}mv\,dv\]
Evaluate integral
\[W_{\mathrm{net}}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2\]
This motivates defining translational kinetic energy from speed, not velocity direction.
Rules
Kinetic energy
\[K=\frac{1}{2}mv^2\]
Work-energy principle
\[W_{\mathrm{net}}=\Delta K=K_f-K_i\]
Final speed
\[v_f=\sqrt{v_i^2+\frac{2W_{\mathrm{net}}}{m}}\]
Examples
Question
A
\[3\,\mathrm{kg}\]
cart moving at \[2\,\mathrm{m\,s^{-1}}\]
has \[48\,\mathrm{J}\]
net work done on it. Find its final speed.Answer
\[K_i=\frac{1}{2}(3)(2^2)=6\,\mathrm{J}\]
\[K_f=6+48=54\,\mathrm{J}\]
\[v_f=\sqrt{\frac{2K_f}{m}}=6\,\mathrm{m\,s^{-1}}\]
Checks
- Kinetic energy is never negative.
- Negative net work reduces speed.
- Work-energy gives speed, not direction.
- Use total work from all forces.