
The Work–Energy Theorem states:
The net work done on an object equals the change in its kinetic energy.
Mathematically:
\(W_{net}=\Delta K\)
where:
Since kinetic energy is:
\(K=\frac{1}{2}mv^2\)
the theorem can be written as:
\(W_{net}=K_f-K_i\)
or
\(W_{net}=\frac{1}{2}mv^2_f-\frac{1}{2}mv^2_i\)
The theorem explains how forces change an object’s motion.
If the net force acts in the direction of motion:
\(W_{net}>0\)
then:
\(K_f>K_i\)
The object speeds up.
Example:
If the net force opposes the motion:
\(W_{net}<0\)
then:
\(K_f<K_i\)
The object slows down.
Example:
If:
\(W_{net}=0\)
then:
\(K_f=K_i\)
The object’s speed remains constant.
Example:
Start with:
\(\sum F=ma\)
Using:
\(a=\frac{dv}{dt}\)
and
\(v=\frac{dx}{dt}\)
we obtain:
\(a=v\frac{dv}{dx}\)
Substitute into Newton’s Second Law:
\(F=mv\frac{dv}{dx}\)
Multiply by dxdxdx:
\(F dx=mv dv\)
Integrate both sides:
\(\int F dx=\int mv dv\)
The left side is work:
\(W_{net}\)
The right side becomes:
\(\frac{1}{2}mv^2_f-\frac{1}{2}mv^2_i\)
Therefore:
\(W_{net}=\Delta K\)
For variable forces:
\(W=\int F(x) dx\)
The work equals the area under the force-versus-position graph.
Once work is found, the Work–Energy Theorem can be used to determine changes in speed.
Using the Work–Energy Theorem often avoids solving for:
This makes many mechanics problems much simpler.
A 2.0 kg object starts from rest.
A net force does 40 (J) of work.
Using:
\(W_{net}=\Delta K\)
\(40=\frac{1}{2}(2)v^2-0\)
\(40=v^2\)
\(v=\sqrt{40}\)
\(v\approx6.3m/s\)
The Work–Energy Theorem is always true.
Energy conservation is a special application that becomes useful when forces are conservative.
The theorem serves as the foundation for later topics such as:
The Work–Energy Theorem states:
\(W_{net}=\Delta K\)
or
\(W_{net}=\frac{1}{2}mv^2_f-\frac{1}{2}mv^2_i\)
Key ideas:
The Work–Energy Theorem provides one of the most powerful problem-solving tools in AP Physics C Mechanics because it directly connects force, work, and motion.
You have not completed all required lessons and assessments.