Work measures energy transfer — For a constant force F acting through displacement s, work is F·s = Fs cos θ, where θ is the angle between them. Work is scalar, measured in joules; the force component parallel to displacement matters. A force perpendicular to motion does no work despite possibly changing its direction.
Positive and negative work — A driving force with a component along displacement does positive work, while resistance opposite motion does negative work. A 20 N force pulling 5 m at 60° to the direction of travel does 20×5×cos 60° = 50 J. Do not use the entire force magnitude when it is oblique.
Variable force and integration — For straight-line motion with position-dependent signed force F(x), work from x = a to b is ∫ₐᵇ F(x) dx. It is signed area under a force–displacement graph. For F = 3x² N and x from 0 to 2 m, work = [x³]₀² = 8 J. A force–time graph instead gives impulse.The schematic shaded area is work because the horizontal axis is displacement. An otherwise similar force–time plot would represent impulse. Exact work for the stated F=3x² example is calculated in the accompanying point; the sketch is not a numerical plot.
Kinetic energy — For a particle of mass m and speed v, K = ½mv². Its nonnegative value depends on speed squared, so opposite velocities of equal magnitude give equal kinetic energies. For a 4 kg particle at 3 m s⁻¹, K = 18 J. Momentum remains a different, directed quantity.
Deriving the work–energy equation — In one dimension, F = ma and a = v dv/dx, so F dx = mv dv. Integrating from u to v gives resultant work = ½mv² − ½mu². Sum work by all forces exactly once. This method can avoid calculating a changing acceleration or the travel time.
A braking calculation — A 1000 kg vehicle slows from 20 to 10 m s⁻¹ under a constant 3000 N resistance alone along a level road. Its kinetic-energy change is ½×1000×(100 − 400) = −150000 J. Thus −3000s = −150000, giving s = 50 m. The negative work represents lost mechanical energy.
Using force as a function of distance — A 2 kg particle starts at 3 m s⁻¹ and gains 7 J net work. Its final kinetic energy is 9 + 7 = 16 J, so v² = 16 and speed is 4 m s⁻¹. Energy determines speed, not the direction of velocity; the path and motion must supply any sign.
Gravitational energy and inclines
Gravitational potential energy — Near Earth's surface with constant g, gravitational potential energy is U = mgh relative to a chosen zero height. Only its change affects energy balances: ΔU = mg(h₂ − h₁). A negative value relative to the chosen datum (zero-height reference) is allowed; it does not mean a negative mass or height difference is impossible.
Gravity and mechanical energy — Gravity's work is −ΔU. You may include this work explicitly in the resultant-work equation or move it into potential energy, but not both. With kinetic plus gravitational potential energy E = K + U, work by all remaining forces equals E_final − E_initial.
Smooth descent — A particle released from rest descends a vertical height h on a fixed smooth track. With no other energy transfer, ½mv² = mgh, giving v = √(2gh). Mass cancels; the energy method does not require a straight track, but contact and the direction of motion must remain physically possible.
Inclined distance versus height — For displacement s up a plane inclined at α to the horizontal, height gain is s sin α and weight's work is −mgs sin α. On a straight smooth plane the normal reaction is perpendicular to displacement and does no work. Using s instead of s sin α overestimates the height unless the plane is vertical.The incline sketch is schematic. Travel distance s is along the plane while height h is vertical. Weight acts downward; the normal reaction is perpendicular to the plane and therefore does no work along a fixed plane. Extra forces and friction must be added for the corresponding models.
Friction on a plane — For a simple particle on a plane with no other perpendicular force, R = mg cos α and limiting/sliding friction in the appropriate model is μR. Work against friction over distance s is μmg cos α × s. An oblique pull can change R; never retain mg cos α without checking the normal balance.
A rough upward journey — A 2 kg particle starts up a 30° plane at 7 m s⁻¹ with μ = 0.2 and g = 9.8. At its first stop, 49 = 2×9.8s(sin 30° + 0.2cos30°), giving s ≈ 3.714 m. Both gravity and friction oppose this upward motion; distinguish this first stop from any subsequent reversal.
Return motion and friction direction — After a turning point, friction reverses with the motion, whereas weight still acts down the plane. Downward motion requires checking whether gravity exceeds available opposing friction; coming to rest does not guarantee immediate sliding back. Write a new directional balance for the return stage rather than extending the old friction sign.
Mechanical energy and variable resistance
Conservative and dissipative forces — A conservative force has work determined only by the initial and final positions. Gravity and ideal elastic forces can be represented by potential energies, so their work depends on endpoints. Friction and resistance generally transfer mechanical energy into other forms. Total energy is conserved overall, but the particle's mechanical energy need not be; define which stores and transfers are included.
A complete balance — A useful form is initial K + initial U + driving work = final K + final U + work against resistance. Every term must describe an energy in joules, and each force or store appears once. Choose initial/final states before substituting numbers, especially when both height and speed change.
Position-dependent resistance — If resistance magnitude is R(x) = 2 + 3x N and motion is in the increasing x direction from 0 to 4 m, work against it is ∫₀⁴(2 + 3x)dx = 8 + 24 = 32 J. The resistance's own signed work is −32 J. Reversing the path requires respecting its opposition to velocity, not blindly keeping a signed force formula.
Velocity-dependent resistance — When resistance is R(v), it cannot generally be integrated with respect to distance as if v were constant. Obtain v(x), use a differential equation, or use a justified supplied relation. Writing R(v)×s without integrating assumes a constant resistance over that travel and may be invalid.
Checking a proposed stopping point — At a proposed stop set K_final = 0 and solve the energy equation. Keep roots compatible with nonnegative travel distance and the stage over which the force law applies. A negative predicted v² at a later position indicates the particle cannot reach that state under the assumed motion, not an imaginary physical speed.
Energy after a change of conditions — Split a journey where a driving force, friction coefficient or track condition changes. Carry the first stage's final speed and height into the next stage. Energy balance is additive, but a force expression valid on one segment must not be applied across the entire route without justification.
Power and driving models
Power is an energy rate — Instantaneous power is dW/dt = F·v. For a force along the motion, P = Fv; units are watts, with 1 W = 1 J s⁻¹. Average power over an interval is total work divided by elapsed time and need not equal the instantaneous value at either endpoint.
Constant power is not constant force — For a vehicle with constant engine power P at positive speed v, driving force is P/v. The force decreases as speed increases. This idealised expression cannot describe starting from exact rest with finite unlimited traction; the model must have a stated valid speed range.
Power and acceleration — Along a road, m dv/dt = P/v − R − mg sin α for an uphill vehicle with resistance R and engine force along the road. At an instant, substitute the current speed to find acceleration. Constant power does not imply constant acceleration because P/v changes with speed.
A level-road example — A 1000 kg car has engine power 20 kW, speed 10 m s⁻¹ and resistance 500 N on a level road. Driving force is 20000/10 = 2000 N, so acceleration is (2000 − 500)/1000 = 1.5 m s⁻². Convert kilowatts to watts before dividing by speed.
Steady climbing — At constant positive speed up an incline, resultant tangential force is zero, so P = (R + mg sin α)v. A 1000 kg car climbing a road with sin α = 0.05 at 10 m s⁻¹ against 500 N resistance requires (500 + 490)×10 = 9900 W. A steady speed is not zero engine output.
Maximum or limiting speed — For positive constant power and constant resistance on a level road, a steady speed satisfies v = P/R. With R depending on v, solve P = R(v)v. For R = kv², v = ∛(P/k). Check the route, signs and whether acceleration approaches the equilibrium from the relevant initial condition.
Energy and power consistency — Multiplying the tangential equation of motion by speed gives dK/dt = driving power − resistance power − gravitational power uphill. This connects Newton's law, work and energy. It also shows why energy in joules cannot be added directly to power in watts: integrate power over time first.
Choice of method and limitations — Use work–energy when displacement, height and endpoint speeds are central; use impulse–momentum for short contact forces or velocity changes; use instantaneous force/power when acceleration is asked at a given speed. State constant mass, fixed track, constant g and resistance assumptions and do not certify a real vehicle's performance from an ideal particle model.
Test yourself
50 questions · Random sets of 10. These quick checks support revision; practise longer explanations and justified judgements too.
Mr Kidman's Classroom
