— Physics · AS Level (7408) · Work Done, Energy Conservation & Power —
★ Exam Tip: "Force × distance" ALONE scores ZERO for "explain
work done" — that's also the formula for torque/moment. You must say "...moved along the line of action of the
force."
1. Work Done
Work done:Force × the distance moved along the line of action of the
force. Scalar, measured in Joules (J). W = F×s
Zero-work example: a person holds a plate with 2N upward force
(cancelling its weight) and walks 5m horizontally. W=0 for that 2N force — in ITS line of action (vertical), the
plate moved no distance at all.
★ Key idea: Work is done whenever energy converts from one form to
another.
Area under an F–s graph = Work done (generalizes W=Fs for a varying force).
Two kinds of work: work done BY a force (same direction as
displacement) vs work done AGAINST a force (opposes displacement — ALWAYS a loss of energy, typically
friction/drag).
Block sliding down a rough inclineWorked example
Block on a rough incline, moving DOWNWARDS. Find the work done by the weight, against friction, and the net
work done.
Work done BY the weight: W = Wsinθ × s (only the parallel component does
work; Wcosθ is perpendicular to the motion, no displacement in that direction)Work done AGAINST friction: W = F_r × s (a loss of energy)Net work done: ΣW = ΣF×s = (Wsinθ − F_r)×s
2. Work Done on/by a Gas (constant P)
Work done ON the gas (compressed)Work done BY the gas (expanding)
Derivation:
W = F×s = P×A×Δl [F=P×A, s=Δl]
A×Δl = ΔV ⟹ W = P × ΔV (P constant)
★ W = PΔV (only valid for CONSTANT pressure)
Area under a P-V graph = Work done (generalizes for varying pressure).
3. Energy & Conservation of Energy
Energy:Ability to do work. Scalar, measured in Joules
(J).
"For a closed system, the total energy remains constant."
★ E_input = E_output + (losses)
★ Scope note: although O-level teaches many energy types (light,
chemical, sound...), only THREE are ever calculated at this level: Gravitational PE, Kinetic energy, Elastic
PE. Others only need basic conceptual understanding.
Types of Potential Energy
Potential energy: Energy an object possesses by virtue of its state (position in a
field, or deformation).
★ Gravitational PE
Energy possessed by virtue of position in a gravitational field (or height above Earth's surface)
★ Elastic PE
Energy possessed by virtue of deformation
Electric PE
Energy possessed by virtue of position in an electric field
GPE: W×s = mg(h₁−h₂)EPE: ½ × F × Δl (area under F-Δl graph)
G.P.E = mgΔh (height taken vertically only) | E.P.E = ½ × F × Δl
Kinetic Energy — Derivation
Kinetic energy: The energy an object possesses by virtue of its motion.
Derivation (via kinematics eqn ③):
W = F×s = m×a×s
v²=u²+2as ⟹ a=(v²−u²)/2s
W = m×[(v²−u²)/2s]×s ⟹ K.E = ½m(v²−u²) ⟹ if u=0: K.E = ½mv²
Pendulum — GPE→KE conversionWorked example
Light inextensible string, length 30cm, swung to 30° from vertical (①, v=0), mass 30g. Find ΔGPE and max speed.
Q1) What energies at ①? G.P.EQ2) ΔG.P.E = mgΔh = (30×10⁻³)(9.81){[30−30cos30]×10⁻²} = 0.0118 ≈ 0.01 JQ3) Max speed: G.P.E=K.E ⟹ 0.01=½(30×10⁻³)v² ⟹ v=√(2×0.01/30×10⁻³) ⟹ v≈0.8 m/s
4. Power
Power: Rate at which work is being done. Scalar, measured in Watts (W). (Also
acceptable: "the rate at which energy is converted from one form to another" — but NOT as the definition of Work
itself.)
① P = dW/dt = ΔW/Δt ⟹ P×Δt = ΔW
Gradient of a W-t graph = Power. Area under a P-t graph = ΔWork done.
② Derivation for constant force:
P = dW/dt = d(F×s)/dt = F×(ds/dt) ⟹ P = F×v (only valid for constant force)
★ P = F × v (constant force only)
Area under an F-v graph = Power (generalizes for varying force).
5. Efficiency
η = (useful energy output)/(total energy input) × 100 | η = (useful power
output)/(total power input) × 100
★ Critical rule: NEVER mix energy and power in the same ratio. Decide
energy-perspective OR power-perspective, keep both top and bottom consistent (convert via E=Pt if given a mix).
★ Self-check: efficiency can NEVER exceed 100%. A result over 100% means
you've made a mistake somewhere.
★ "Nothing in the universe is 100% efficient." (Also: a system's total energy content can never exceed what was
originally put in — useful for debunking "free energy machine" claims in theory questions.)
6. Worked MCQ Practice — Equilibrium ⟹ P=Fv
★ Reusable method: constant velocity ⟹ EQUILIBRIUM ⟹ the
driving/thrust force exactly balances whatever opposes it (weight component, drag, resistive force). Find that
balancing force, then apply P=Fv directly.
Escalator lifting peopleMCQ
Escalator: 250 people (78kg each) at constant 4.3 m/s, angle 40° to horizontal. Minimum power required?
Constant velocity ⟹ equilibrium: driving force F_T = mgsinθ (only the
parallel weight component needs balancing — mgcosθ cancels with R, no work done)P = F_T×v = (250×78)×9.81×sin40×4.3 ≈ 530 kW
Answer: C (530 kW)
Aircraft in horizontal flightMCQ
Aircraft at constant 90 m/s, drag=2400N, mass 2000kg. Power produced by the thrust force?
Car travelling up a slope, same constant speedMCQ · two-stage
equilibrium
Car 1400kg at constant 25 m/s on flat road, output power 30kW. Car then climbs a 2° slope at the SAME constant
speed. New output power?
Flat road (equilibrium): P=F_r×v ⟹ 30000=F_r×25 ⟹ F_r=1200N (resistive
force, assumed unchanged on the slope)On slope: new driving force = resistive force + weight component: F_T2 = mgsinθ + F_r =
1400×9.81×sin2° + 1200P = F_T2 × v = (1400×9.81×sin2+1200)×25
Answer: C (42 kW)
7. Structured Problems
★ Major strategic insight: whenever a question involves
NON-negligible air resistance (drag not constant), kinematics equations (constant a) AND F=ma both become unusable
for exact values — that's why such questions ask "qualitatively" not "quantitatively" in Kinematics/Forces.
Energy conservation is the workaround: total energy is ALWAYS conserved regardless of how acceleration
varies, so you CAN find an AVERAGE resistive force via E_input=E_output even when F=ma can't help. Recognise
"significant air resistance + find a force" as the cue to reach for energy methods.
Cyclist up a slope — v-t graph, full energy balanceStructured
Q
Cyclist+bicycle (92kg combined) climbs a slope in 8.0s; v-t graph given; vertical height climbed 1.3m.
(a) Distance up the slope = area under the v-t graph (count boxes) = 39
m(b)(i)1. ΔKE = ½m(v²−u²) (reading v≈3, u≈6.8 from graph) = ½(92)(3²−6.8²) ⟹ −1240 J (a
loss)(b)(i)2. ΔGPE = mgΔh = (92)(9.81)(1.3) ⟹ 1170 J(b)(ii) Useful power 75W: P=ΔW/Δt ⟹ 75×8=ΔW ⟹ ΔW=600 J (work done by the cyclist
pedalling)
Bookkeeping convention: a DECREASE in some energy (KE lost) is an
INPUT to the rest of the system; an INCREASE (GPE gained) is an OUTPUT; work actively added (pedalling) is an
INPUT. Energy is scalar — don't carry the negative sign into the equation, just use it to sort input vs output.
(d) Why isn't the resistive force constant? "The velocity is decreasing and so the drag force also
decreases."
Derive GPE, then water wheel (efficiency + kinematics)Structured
Q
(b) Derive GPE=mgh from the defining equation for work done: W=F×s ⟹ W×h ⟹ mgh=G.P.E
(c) Water falls 120m from a reservoir to a water wheel. At the wheel, KE of the water is only 10%
of its GPE at the reservoir.
(i) Show v=15 m/s: KE=(10/100)×GPE ⟹ ½mv²=(1/10)(m)(9.81)(120) — mass
cancels ⟹ v²=(2/10)(9.81)(120)=235.44 ⟹ v≈15.3≈15 m/s(ii) Water wheel produces 110kW electrical power, conversion 25% efficient. Find mass of
water/second:η=(useful power output)/(total power input)×100: 25 = [110×10³]/[½(m/t)v²] ×100m/t = (110×10³×100)/(½×235.44×25) ⟹ ≈3700 kg s⁻¹
Trick: mass wasn't given directly for (i) — it cancelled from
both sides of the GPE=KE equation. Always try proceeding symbolically before assuming a question is unsolvable
due to "missing" data.
Leisure-park carriage — energy + momentum combinedStructured Q —
multi-topic
Carriage 580kg follows a quarter-circle arc XY, radius 13m. At X: 22 m/s horizontal. At Y: 12 m/s vertical.
★ Common mistake: when sketching a-t from 0 to T, if the ball
reaches terminal velocity BEFORE hitting the ground, the curve must flatten to EXACTLY ZERO and stay flat for a
visible stretch before impact — don't just let it curve asymptotically all the way to t=T.
a: starts at g, decays, flattens to exactly 0 before t=T
Full energy-conversion explanation (before/after terminal
velocity):Before terminal velocity: ↓GPE → KE↑ + W_Fr(heat) — GPE converts into BOTH increasing KE AND heat
(friction/drag), simultaneously. After terminal velocity (constant speed): ΔKE=0 exactly (v not changing) ⟹ ALL lost GPE converts into
heat only: ↓GPE → W_Fr(heat).
✓ Quick Recap — Full Chapter Cheat-Sheet
W = F×s along the line of action — "along the line of action" is mandatory wording, or zero marks.