Work, Energy and Power — Explained
Detailed Explanation
Work is an energy transfer with a direction condition
In mechanics, work done by a force depends on both the force and the displacement of its point of application. For a constant force acting on a particle, W = F s cos θ, where θ is the angle between the force and displacement. The cosine matters: a large force does not imply large work if it is perpendicular to the displacement. The quantity is a scalar, although it is calculated from vectors. Its SI unit is the joule, equivalent to a newton metre.
Suppose a constant 30-newton force pulls an object through 4 metres at an angle of 60 degrees to its displacement. Its work is 30×4×0.5=60 joules, not 120. The latter figure uses the entire force as though it acted along the displacement.
This example specifies the geometry and assumes the force remains constant; it is not a measurement of a particular machine. If force varies, the calculation must account for the force over the path rather than multiplying one arbitrary force reading by the complete distance.
The sign belongs to the force being considered
A force component along displacement does positive work; an opposing component does negative work. Consider a block sliding two metres while a constant friction force of five newtons opposes its motion. Friction does −10 joules of work on the block. The negative sign does not mean that energy ceases to exist. It indicates how that force affects the block's mechanical-energy account, with energy transferred into internal forms in the wider system.
Several forces can act on the same object. An applied force might do positive work while friction does negative work and the normal force does zero work along an ideal horizontal displacement. Do not identify one force's contribution with net work. A person holding a bag stationary may become tired, but the bag has no displacement in that ideal description. Zero mechanical work on the stationary bag is not a claim of zero physiological energy use by the person.
Net work determines the change in kinetic energy
For a particle of constant mass in ordinary nonrelativistic mechanics, kinetic energy is K = one-half m v squared. The work–energy theorem states that net work equals the change in kinetic energy. A two-kilogram cart speeding up from three to five metres per second gains one-half×2×(25−9)=16 joules. Subtracting the speeds before squaring would instead produce four joules, which is not the difference between the two kinetic energies.
The theorem connects initial and final speeds without requiring the time history of the motion. It does not say that any one applied force supplies the whole change when other forces do work. If the pull contributes 30 joules and friction contributes −14 joules, the net contribution is 16 joules. Those numbers are a hypothetical energy balance. An observed final speed would need the actual forces, initial state and model conditions before being attributed to one cause.
Potential energy belongs to a specified interaction
Near Earth's surface, for a sufficiently small height change that gravitational acceleration can be treated as constant, the change in gravitational potential energy is m g Δh. Raising three kilograms through two metres with g taken as ten metres per second squared increases that energy by 60 joules. The assumed g is a rounded exercise value, not a universal exact constant. The zero of potential energy can be chosen conveniently; differences govern the energy changes in this calculation.
For an ideal spring obeying Hooke's law, stored elastic potential energy is one-half k x squared relative to the unstretched state. Doubling the extension therefore multiplies the stored energy by four within that model. It does not merely double it. The force also changes during extension, so multiplying the final spring force by the whole extension would overstate the work. The spring relation must not be extended beyond the elastic behaviour assumed by the model.
Conservation requires the correct system boundary
When only conservative forces perform work within the chosen mechanical account, the sum of kinetic and potential energy remains constant. In an ideal fall without air resistance, a loss of gravitational potential energy corresponds to a gain of kinetic energy. With drag, part of the energy is transferred to internal energy of the surroundings; the object's mechanical energy is not conserved even though a wider total-energy account remains valid.
This distinction prevents a misleading phrase such as energy being lost from becoming a claim of destruction. Specify lost from which useful store or system. A machine may transfer only part of its input into the desired output, with the remainder leaving as other forms. Its efficiency compares useful output with input under a defined boundary. Two devices cannot be compared fairly if one figure includes auxiliary consumption and the other excludes it.
Power measures rate, not a larger kind of energy
Average power is transferred energy or work divided by elapsed time. Two hoists that each deliver 600 joules but take three and six seconds have average useful powers of 200 and 100 watts. They deliver the same useful energy at different rates. Their electrical consumption cannot be inferred from these useful-output figures alone without efficiency or input measurements. A power rating is also not proof that the device operates continuously at that rate.
A kilowatt-hour is an energy unit: 1,000 joules per second multiplied by 3,600 seconds equals 3.6 million joules. It is not interchangeable with a kilowatt. A 0.8-kilowatt appliance operating at that input for 2.5 hours consumes two kilowatt-hours, or 7.2 million joules. Keep the duration and the assumption of constant input explicit. A clear answer connects the formula to its physical meaning, checks the units and signs, and names the system whose energy is being counted.