Work-Energy Theorem Examples in Physics
Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Work-Energy Theorem.
This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Physics.
Concept Recap
The net work done on an object by all forces acting on it equals the change in its kinetic energy.
The total work done on an object is exactly what changes its kinetic energy.
Read the full concept explanation →How to Use These Examples
- Read the first worked example with the solution open so the structure is clear.
- Try the practice problems before revealing each solution.
- Use the related concepts and background knowledge badges if you feel stuck.
What to Focus On
Core idea: This theorem bridges force-based thinking (work) with energy-based thinking (kinetic energy).
Common stuck point: Only net work changes KE—individual forces may do positive or negative work.
Sense of Study hint: When using the work-energy theorem, first identify all forces doing work on the object. Then calculate the net work: add positive work (forces in the direction of motion) and subtract negative work (forces opposing motion). Finally, set W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 and solve for the unknown.
Worked Examples
Example 1
easySolution
- 1 The work-energy theorem states: W_{\text{net}} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2.
- 2 Net work done: W = Fd = 20 \times 5 = 100 \text{ J}.
- 3 Since v_i = 0: 100 = \frac{1}{2}(4)v_f^2 \implies v_f = \sqrt{\frac{200}{4}} = \sqrt{50} \approx 7.07 \text{ m/s}
Answer
Example 2
mediumPractice Problems
Try these problems on your own first, then open the solution to compare your method.
Example 1
mediumExample 2
hardRelated Concepts
Background Knowledge
These ideas may be useful before you work through the harder examples.