M2.3 - Work and energy

Syllabus
2019
Topic
M2.3
Level
A2

Track work, energy and power through motion

Energy methods compare two states without requiring the time taken between them. Kinetic energy depends on speed, gravitational potential energy depends on vertical height, and work transfers energy when a force acts through a displacement.

K=12mv2,U=mgh,W=FscosϕK=\frac12mv^2,\qquad U=mgh,\qquad W=Fs\cos\phi

Here mm is mass, vv is speed, hh is height above any consistent datum, and ϕ\phi is the angle between a constant force and the displacement. Energy and work are measured in joules. Only changes in potential energy matter, so choose a convenient zero height and preserve the sign of Δh\Delta h.

Kf+Uf=Ki+Ui+WotherK_f+U_f=K_i+U_i+W_{\rm other}

Contribution Entry in the energy ledger
gravity represented by the change ΔU=mgΔh\Delta U=mg\Delta h
driving or pushing force along motion positive work +Fs+Fs
constant resistance or friction opposite motion negative work Rs-Rs
normal reaction perpendicular to motion zero work
no non-conservative work K+UK+U is conserved

The work-energy principle is equivalently ΔK=\Delta K= total work done by all forces. Using gravitational potential energy in the ledger avoids also counting gravity as work: choose one representation, not both. On an incline of length ss and angle θ\theta, the vertical change is ssinθs\sin\theta, while a constant resistance of magnitude RR dissipates energy RsRs in either direction of travel.

A 44 kg particle moves 55 m up an incline whose sine is 0.30.3. Its initial speed is 8ms18\,\mathrm{m\,s^{-1}} and a constant resistance of 66 N opposes motion. With no driving work,12(4)v2+4g(5×0.3)=12(4)(82)6(5).\frac12(4)v^2+4g(5\times0.3)=\frac12(4)(8^2)-6(5).Thus 2v2+58.8=982v^2+58.8=98, so v=4.43ms1v=4.43\,\mathrm{m\,s^{-1}}. The calculation accounts separately for increased potential energy and energy dissipated by resistance.

P=dWdt=Fvwhen the force is parallel to the velocityP=\frac{dW}{dt}=Fv\quad\text{when the force is parallel to the velocity}

Power is the rate of doing work, measured in watts, with 1kW=1000W1\,\mathrm{kW}=1000\,\mathrm W. At a particular speed, a parallel driving force is F=P/vF=P/v; therefore constant power produces a smaller driving force at a larger speed. For example, 1212 kW at 20ms120\,\mathrm{m\,s^{-1}} corresponds to 600600 N before resistance and weight components are included in F=maF=ma.

Mechanical energy is not conserved when driving work or resistance is present, although the full energy ledger still balances. Do not use signed velocity in 12mv2\tfrac12mv^2, count gravitational work and mghmgh twice, use slope distance as vertical height, or treat constant power as constant force. Collision-specific energy loss belongs to the next Topic.