Work, energy and power flashcards

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Kinetic energy
The energy a body has because of its motion: KE = ½mv², in joules. A scalar, never negative.
Gravitational potential energy
The energy a body has because of its height above a chosen zero level: GPE = mgh, where h is the vertical height.
Work done by a constant force
The magnitude of the force × the distance moved in the direction of the force: W = Fs cos θ.
The joule
The work done when a force of 1 N moves its point of application 1 m in the direction of the force. 1 J = 1 N m.
Power
The rate at which work is done. Unit the watt: 1 W = 1 J s⁻¹.
Work–energy principle
The change in the total mechanical energy (KE + PE) of a body equals the work done on it by forces other than its weight.
Conservation of mechanical energy
If the only force doing work on a body is its weight, KE + PE stays constant.
Maximum (limiting) speed of a vehicle at constant power
The speed at which the acceleration is zero: the driving force P/v equals the total resistance (plus mg sin α uphill).
Formula for kinetic energy
KE = ½mv² (No — must be known)
Formula for gravitational potential energy
GPE = mgh (h vertical, g = 9.8 m s⁻²) (No — must be known)
Work done by a force F at angle θ to the motion over a distance s
W = Fs cos θ (No — derive it)
Power from force and speed
P = Fv (No — derive it)
Driving force of an engine working at constant power P at speed v
F = P/v (No — derive it)
Newton's second law for a vehicle at constant power moving up a slope
P/v − R − mg sin α = ma (No — derive it)
Work–energy equation
KE₁ + PE₁ + work done by driving forces = KE₂ + PE₂ + work done against resistances (No — derive it)
Graph of driving force against speed for an engine at constant power
A curve F = P/v (a reciprocal graph): the driving force falls as speed rises. Maximum speed is where it meets the resistance line.
KE and GPE against distance fallen for a particle moving freely under gravity
GPE falls in a straight line, KE rises in a straight line with equal and opposite gradient; their sum is a horizontal line.
Energy bar chart for a particle sliding down a rough slope from rest
PE lost = KE gained + work done against friction: the KE bar is shorter than the PE bar by the friction work.
Forces to show along the slope for a car driving uphill
Driving force up the slope; resistance and mg sin α down the slope. R and mg cos α are perpendicular and balance.
Shape of the v–t graph of a car starting from rest with its engine at constant power
A curve with decreasing gradient that levels off towards the maximum speed (the acceleration falls as P/v falls).
Why does the normal reaction do no work on a body sliding along a surface?
It is perpendicular to the direction of motion, so its component along the motion is zero (cos 90° = 0).
Why does the acceleration of a car at constant power decrease as it speeds up?
The driving force P/v decreases as v increases, while the resistance stays the same, so the resultant force decreases.
Why does the speed of a projectile at a given height not depend on its angle of projection?
KE + PE is conserved and depends only on speed and height, so the same height gives the same speed whatever the direction.
When can you NOT use conservation of mechanical energy?
When friction, resistance or a driving force does work. Use the work–energy principle with the extra work terms.
Why can't the constant-acceleration formulae be used for a vehicle at constant power?
The driving force P/v changes with speed, so the acceleration is not constant. Use energy or Newton's second law at an instant.
Kinetic energy of a 900 kg car at 20 m s⁻¹?
½ × 900 × 20² = 180 000 J = 180 kJ.
A 50 kg runner goes 300 m up a road with sin α = 1/25. Gain in GPE?
Height = 300/25 = 12 m; GPE = 50 × 9.8 × 12 = 5880 J.
Work done by a 40 N force at 30° to the motion over 25 m?
40 cos 30° × 25 = 866 J (3 s.f.).
A van works at 15 kW against a constant 600 N resistance on a level road. Maximum speed?
15 000 / v = 600 → v = 25 m s⁻¹.
An 800 kg car at 20 kW, resistance 600 N, level road. Acceleration at 10 m s⁻¹?
Driving force 2000 N; 2000 − 600 = 800a → a = 1.75 m s⁻².
A particle slides from rest down a smooth slope through a vertical drop of 5 m. Speed at the bottom?
½v² = 9.8 × 5 → v = 9.90 m s⁻¹ (3 s.f.).
A 60 kg sledge slows from 8 to 4 m s⁻¹ over 40 m of level snow. Constant resistance?
KE lost = ½ × 60 × (64 − 16) = 1440 J; R = 1440 / 40 = 36 N.
Change in KE written as ½m(v − u)²
Wrong: change in KE = ½mv² − ½mu². Square each speed first.
Using the distance along a slope as h in mgh
h is the vertical height: h = d sin α.
Power given in kW
Convert to watts before using F = P/v: 24 kW at 12 m s⁻¹ gives 2000 N.
Question says “use the work–energy principle”
Write an energy equation in words, then substitute. A solution by F = ma and suvat does not follow the instruction.
Gravity counted twice
Include the weight either as a PE change or as a work term, never both.
Answer accuracy with g = 9.8
Give final answers to 2 or 3 significant figures and keep full values in the working.

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