Work, Energy and Power · Lesson 6 of 16
Conservation of Mechanical Energy
“Derive mechanical-energy conservation and apply it to falling bodies, slopes and pendulums.”
• Derive conservation from the work–energy theorem. • Choose the system and zero level. • Calculate speeds from height changes and pendulum geometry.
Combine kinetic and potential energy
For a particle acted on by a conservative force, the work–energy theorem says ΔK = Wc. Potential energy gives Wc = −ΔU. Combining these statements gives ΔK + ΔU = 0. Thus kinetic and potential energies may each change while their sum remains fixed. Forces such as an ideal stationary track’s normal reaction or a fixed-length pendulum’s tension can be present without spoiling this result if they do zero work.
Draw the body and name its interactions before cancelling any work. For gravity, the Earth–body system supplies the stored gravitational energy. For a spring, include its elastic energy. A motor can add energy and friction can transfer organised mechanical energy into internal energy, so the mechanical sum is not automatically conserved in every problem. Total energy has a broader meaning than mechanical energy.
Free fall, at every height
At the release point all the mechanical energy is potential. At intermediate heights some is kinetic. Just above the ground it is entirely kinetic under this reference choice. Mass cancels from the speed relation because both the energy released and the kinetic energy at a given speed are proportional to mass. A nonzero initial speed contributes an initial kinetic-energy term and must not be omitted.
A smooth slope changes timing, not the height result
Problem
Two beads start from rest at the same height 5 m above a common endpoint and follow different stationary smooth tracks. Use g = 10 m/s². Compare final speeds.
- 1.The normal force is perpendicular to motion and does zero work. Gravity work depends only on vertical drop.
- 2.For either bead, ½mv² = mg×5, giving v = 10 m/s.
- 3.Different path lengths and accelerations can produce different travel times. Equal final speeds do not mean equal trajectories or times.
Pendulum height from geometry
A bob on a light inextensible string moves on a circle about a fixed pivot while the string is taut. At angle θ from the downward vertical, its vertical distance below the pivot is L cosθ. It is therefore L − L cosθ above the lowest point. Tension is radial and instantaneous motion tangential, so tension does no work in this ideal model. Gravity changes the speed.
Problem
A pendulum of length 2 m is released from rest at 60° to the vertical. Use g = 10 m/s². Find bottom speed.
- 1.The release height above the bottom is 2(1 − 0.5) = 1 m.
- 2.Energy conservation gives ½mv² = mg×1.
- 3.The bottom speed is √20 ≈ 4.47 m/s. Tension changes along the arc but contributes no work while the pivot and length are fixed.
Carrying a load versus lifting it by a pulley
Problem
A person walks 2 m holding a 15 kg load at constant height, then consider walking 2 m while pulling a rope over an ideal fixed pulley to lift a 15 kg mass. Use g = 9.8 m/s².
- 1.In the first idealisation, the support force on the load is vertical while its displacement is horizontal. The person does zero work on the load through this force.
- 2.With a taut inextensible rope over one fixed pulley, pulling 2 m lifts the mass 2 m. At steady speed the tension is mg.
- 3.The person’s work on the lifted mass is mgh = 15×9.8×2 = 294 J. This is larger than zero even though the person walks the same horizontal distance.
- 4.Human metabolic expenditure includes internal processes and is not equal to the load’s mechanical work.
Check your understanding
Quiz
Mechanical energy is
Explanation: It combines kinetic energy and the included interaction potential energies.
On a fixed smooth track, normal reaction usually does
Explanation: Its direction is perpendicular to the local motion.
Two frictionless paths with the same vertical drop give the same
Explanation: Energy depends on height difference, while timing depends on the path.
Pendulum height above the bottom is
Explanation: Subtract the vertical pivot-to-bob distance from L.
A falling body loses 30 J of U without other work. Its K changes by
Explanation: ΔK = −ΔU.
Key Takeaways
• Derive K+U conservation by combining two work statements. • Height differences determine gravity energy changes. • Apply geometry and force constraints before using an energy result.