The Work-Kinetic Energy Theorem states that the net work done by all forces on an object equals the change
in its kinetic energy. It is useful because it connects forces and displacement directly to speed, without
requiring a full time-based motion calculation.
Main theorem
\[
W_{\mathrm{net}}=\Delta K=K_f-K_0
\]
\[
K=\frac12mv^2
\]
Therefore:
\[
W_{\mathrm{net}}=\frac12mv_f^2-\frac12mv_0^2
\]
Constant net force over a straight displacement
If the net force is constant and points along the line of motion:
\[
W_{\mathrm{net}}=F_{\mathrm{net}}s
\]
Combining this with the theorem gives:
\[
F_{\mathrm{net}}s=\frac12mv_f^2-\frac12mv_0^2
\]
Solving for final speed:
\[
v_f=\sqrt{v_0^2+\frac{2F_{\mathrm{net}}s}{m}}
\]
Acceleration connection
If the acceleration is constant:
\[
F_{\mathrm{net}}=ma
\]
\[
W_{\mathrm{net}}=mas
\]
This agrees with the kinematic equation:
\[
v_f^2=v_0^2+2as
\]
Friction and applied forces
On a horizontal surface with an angled applied force:
\[
F_{\parallel}=F\cos\theta
\]
If the force has an upward component, the normal force is:
\[
N=mg-F\sin\theta
\]
Kinetic friction has magnitude:
\[
f_k=\mu_kN
\]
Since kinetic friction opposes the motion, its work is negative:
\[
W_f=-f_ks
\]
The net work can be written as:
\[
W_{\mathrm{net}}=W_F+W_f+W_{\mathrm{other}}
\]
\[
W_{\mathrm{net}}=(F\cos\theta)s-(\mu_kN)s+F_{\mathrm{other}}s
\]
Sign of net work
| Condition |
Energy effect |
Motion effect |
| \(W_{\mathrm{net}}>0\) |
\(\Delta K>0\) |
The object speeds up. |
| \(W_{\mathrm{net}}<0\) |
\(\Delta K<0\) |
The object slows down. |
| \(W_{\mathrm{net}}=0\) |
\(\Delta K=0\) |
The speed stays the same. |
Stopping before the requested distance
If the equation gives a negative final kinetic energy,
\[
K_f=K_0+W_{\mathrm{net}}<0,
\]
then the object cannot physically complete the whole displacement under that negative net work. It stops first.
For a constant opposing net force, the stopping distance is:
\[
s_{\mathrm{stop}}=\frac{K_0}{-F_{\mathrm{net}}}
\]
Worked example
A block of mass \(8\ \mathrm{kg}\) is pushed by a horizontal force of \(45\ \mathrm{N}\) over \(12\ \mathrm{m}\).
The kinetic friction coefficient is \(\mu_k=0.30\), and the block starts from rest.
Friction force.
\[
N=mg=(8)(9.81)=78.48\ \mathrm{N}
\]
\[
f_k=\mu_kN=(0.30)(78.48)=23.54\ \mathrm{N}
\]
Net force and net work.
\[
F_{\mathrm{net}}=45-23.54=21.46\ \mathrm{N}
\]
\[
W_{\mathrm{net}}=F_{\mathrm{net}}s=(21.46)(12)=257.5\ \mathrm{J}
\]
Final speed.
\[
\frac12mv_f^2=257.5
\]
\[
v_f=\sqrt{\frac{2(257.5)}{8}}\approx 8.02\ \mathrm{m/s}
\]
Formula summary
| Concept |
Formula |
Meaning |
| Work-Kinetic Energy Theorem |
\(W_{\mathrm{net}}=\Delta K\) |
Net work equals change in kinetic energy. |
| Kinetic energy |
\(K=\frac12mv^2\) |
Energy of motion. |
| Constant net force work |
\(W_{\mathrm{net}}=F_{\mathrm{net}}s\) |
Work from constant force along displacement. |
| Final speed |
\(v_f=\sqrt{v_0^2+\frac{2W_{\mathrm{net}}}{m}}\) |
Speed from net work. |
| Acceleration from work |
\(a=\frac{W_{\mathrm{net}}}{ms}\) |
Equivalent constant acceleration. |
| Kinetic friction work |
\(W_f=-\mu_kNs\) |
Friction removes mechanical energy. |
The key idea is energy bookkeeping: initial kinetic energy plus net work becomes final kinetic energy.
