The work–energy principle says that the work done by all forces on a particle equals its change in kinetic energy: T₁ + ΣU₁₋₂ = T₂. It connects speed to position directly, so problems that ask for a speed after some distance, or a distance needed to stop, can be solved without finding the acceleration or the time.
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Where the principle comes from
Start from Newton’s second law along the path, ΣFₜ = m·aₜ, and use the kinematics relation aₜ·ds = v·dv. Integrating from position 1 to position 2 gives ∫ΣFₜ·ds = ½·m·v₂² − ½·m·v₁². The left side is the work done by the forces; the right side is the change in kinetic energy T = ½·m·v².
Work and energy are scalars, so there are no components to resolve. That is the method’s main advantage over ΣF = m·a.
The work of common forces
Only force components along the motion do work. A normal force on a fixed surface does none. Gravity does −m·g·Δy, which depends only on the change in height, not the path. A linear spring does −½·k·(s₂² − s₁²), measured from its unstretched length. Kinetic friction does −μₖ·N·d, always negative, and turns mechanical energy into heat.
Draw a free body diagram first, exactly as in statics. Every force on it either does work or does not; list them one by one.
Conservation of energy and power
When only gravity and springs do work, both can be written as potential energies, V_g = m·g·y and V_e = ½·k·s², and the principle becomes T₁ + V₁ = T₂ + V₂. Friction or a motor breaks that balance, so keep them as work terms.
Power is the rate of doing work, P = F·v. Efficiency is output power over input power, which is always below one for a real machine.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Principle of work and energy
T₁ + ΣU₁₋₂ = T₂, T = ½·m·v²
m = mass (kg), v = speed (m/s), U and T in joules
Hibbeler, Engineering Mechanics: Dynamics
Work of gravity
U = −m·g·(y₂ − y₁)
y measured upward (m), g = 9.81 m/s²
Beer et al., Vector Mechanics for Engineers: Dynamics
Work of a linear spring
U = −½·k·(s₂² − s₁²)
k = stiffness (N/m), s = deformation from unstretched length (m)
Beer et al., Vector Mechanics for Engineers: Dynamics
Work of kinetic friction
U = −μₖ·N·d
μₖ = kinetic friction coefficient, N = normal force (N), d = sliding distance (m)
Meriam & Kraige, Engineering Mechanics: Dynamics
Power and efficiency
P = F·v, ε = P_out / P_in
P in watts (W = J/s)
Hibbeler, Engineering Mechanics: Dynamics
Worked examples
Example: A tote sliding down a gravity chute
Given
A 25 kg tote enters a straight chute 4.0 m long, inclined at 20°, at 0.5 m/s.
Take μₖ = 0.15 between the tote and the chute liner, an assumed value for illustration.
Find
The speed of the tote at the bottom of the chute.
Solution
Initial kinetic energy.T₁ = ½ × 25 × 0.5² = 3.1 J
Work of gravity over a 4.0 m slide, with a drop of 4.0·sin 20°.U_g = 25 × 9.81 × 4.0 × sin 20° = 335.5 J
The normal force balances the weight component perpendicular to the chute, so friction does negative work.N = 25 × 9.81 × cos 20° = 230.5 N, U_f = −0.15 × 230.5 × 4.0 = −138.3 J
Apply T₁ + ΣU = T₂. The normal force does no work.T₂ = 3.1 + 335.5 − 138.3 = 200.4 J
Solve for speed.v₂ = √(2 × 200.4 / 25) = 4.00 m/s
Answer: The tote reaches about 4.0 m/s at the bottom. Friction removes about 41% of the energy gravity supplies, so a small change in μₖ changes the exit speed noticeably.
Example: Sizing a spring buffer for a rail trolley
Given
A 400 kg trolley runs into an end-of-travel spring buffer at 1.2 m/s.
The buffer is a linear spring with k = 60 kN/m. Rolling friction is neglected.
Find
The maximum compression of the buffer.
The peak force and deceleration.
Solution
Kinetic energy at contact.T₁ = ½ × 400 × 1.2² = 288 J
At maximum compression the trolley stops, so T₂ = 0, and the spring has absorbed all of T₁.288 − ½ × 60 000 × s² = 0
Solve for the compression.s = √(2 × 288 / 60 000) = 0.098 m
Peak spring force, and the deceleration it causes.F = 60 000 × 0.098 = 5880 N, a = 5880 / 400 = 14.7 m/s² ≈ 1.5 g
Answer: The buffer compresses about 98 mm and pushes back with a peak of about 5.9 kN, a deceleration near 1.5 g. A softer spring lowers the force but needs more travel; the energy, 288 J, is fixed by the mass and speed.
Common mistakes
Adding work as vectors. Work and kinetic energy are scalars. Add them with signs, not components.
Measuring spring deformation from the wrong point. s is the stretch or compression from the spring’s unstretched length, not from where the motion starts.
Counting gravity twice. Include gravity either as work U_g or as potential energy V_g, never both.
Using mass instead of weight for friction. Friction is μₖ·N, and N is a force in newtons. On a slope, N = m·g·cos θ, not m·g.
Using energy methods to find time. Work–energy gives speed versus position. If the question asks how long, use kinematics or impulse and momentum.
In industry
Frequently asked questions
Is the work–energy principle the same as conservation of energy?
Conservation of energy, T₁ + V₁ = T₂ + V₂, is the special case where only conservative forces such as gravity and springs do work. The work–energy principle also handles friction and applied forces.
Does the normal force ever do work?
Not on a fixed surface, because it is perpendicular to the motion. It can do work when the surface itself moves, such as the floor of an accelerating lift.
When should I use work–energy instead of F = m·a?
When the question links speed and position and the forces are known functions of position. If it asks for time or an acceleration at an instant, F = m·a is usually more direct.
Can I use it for a system of particles?
Yes. Add the kinetic energies of all particles and the work of all external forces, plus the work of internal forces that do not cancel, such as friction between parts that slide on each other.