Work, Energy and Power · Lesson 3 of 16
Kinetic Energy and the Work–Energy Theorem
“Derive the link between net work and speed, then use it in friction and stopping problems.”
• Derive the theorem for constant force. • Add the work of all forces. • Relate speed, kinetic energy and momentum.
Energy because of motion
A moving ball can compress a spring; flowing water can turn a wheel. Translational kinetic energy measures the work available from a particle’s motion. It is proportional to mass and to the square of speed. Doubling speed multiplies kinetic energy by four. Reversing velocity without changing speed leaves kinetic energy unchanged, even though momentum reverses.
Kinetic energy is nonnegative and scalar. Its value depends on the reference frame because speed does. A passenger at rest relative to a train has zero translational kinetic energy in the train frame and nonzero energy in the ground frame. State the frame and use it consistently for the forces, work and velocities.
Derive the constant-force theorem
For constant acceleration, the vector kinematic relation is v² − u² = 2a·d. Multiply by m/2, then use Newton’s second law ma = Fnet. The left side becomes the change in kinetic energy and the right side becomes the dot product of the resultant force with displacement.
A positive net work increases speed; a negative value decreases speed. Zero net work between two positions means equal endpoint speeds, not necessarily constant speed at every intermediate point. The theorem is derived from Newton’s second law; it is another form of the same dynamics, not a separate force law. It normally determines speed, not the direction of velocity or time taken.
Problem
A 2 kg block starts from rest under a 7 N horizontal pull. Kinetic friction coefficient is 0.1. Find the work terms in 10 s, using g = 9.8 m/s².
- 1.Normal reaction is mg = 19.6 N, so friction is 1.96 N. Net horizontal force is 5.04 N and acceleration is 2.52 m/s².
- 2.Displacement in 10 s is ½×2.52×100 = 126 m.
- 3.Pull work = 7×126 = 882 J. Friction work = −1.96×126 = −246.96 J. Normal and weight do zero work.
- 4.Net work = 635.04 J. Final speed is 25.2 m/s, giving ΔK = ½×2×25.2² = 635.04 J, as required.
Problem
A 1.00 g drop falls from rest through 1.00 km and reaches 50.0 m/s. Take g = 10 m/s². Calculate gravity and drag work.
- 1.Gravity work is mgh = 0.001×10×1000 = 10.0 J.
- 2.The final kinetic energy is ½×0.001×50² = 1.25 J.
- 3.The theorem gives Wdrag = ΔK − Wgravity = 1.25 − 10.0 = −8.75 J.
- 4.The detailed drag law is unnecessary for its total work. Do not equate gravity work alone to ΔK while drag is present.
Stopping distance and energy fractions
Problem
A 0.050 kg bullet enters plywood at 200 m/s and emerges with 10% of its initial kinetic energy. Find exit speed.
- 1.Initial kinetic energy is 1000 J; final kinetic energy is 100 J.
- 2.Since K is proportional to v² for fixed mass, vf/vi = √0.10.
- 3.The exit speed is 200√0.10 = 63.25 m/s, a speed reduction of about 68.4%, not 90%.
Check your understanding
Quiz
If speed doubles at fixed mass, kinetic energy becomes
Explanation: K is proportional to speed squared.
Which work equals ΔK?
Explanation: All forces must be included.
Equal initial and final speeds imply
Explanation: The endpoint kinetic energies are equal.
At equal momentum magnitude, the lighter particle has
Explanation: K = p²/(2m).
A bullet retaining 25% of its K retains what fraction of speed?
Explanation: The speed ratio is the square root of the energy ratio.
Key Takeaways
• Net work, rather than one selected work, controls kinetic-energy change. • Kinetic energy is scalar, frame dependent and quadratic in speed. • Energy methods often avoid finding the full force history.