Work, energy and power

Work depends on the component of force along the displacement: W = Fs cos θ for a constant force. A support can exert a large force while a held object remains still, yet transfer no mechanical energy to that object through movement. The person holding it may nevertheless consume metabolic energy internally. In uniform circular motion, the inward force is perpendicular to velocity and therefore does no work, although it continuously changes velocity direction. Do not extend this argument to a moving support or a force whose point of application moves along its own direction.

The derivation uses net work, because it is the resultant force that determines the acceleration. If resistance acts, work done by a pulling force alone is larger than the kinetic-energy gain. The algebra also makes clear that a nonzero starting velocity should not simply be dropped: the energy change depends on the difference of the squared speeds. This classical expression assumes a non-relativistic object of constant mass. A dimensional check confirms units of joules but cannot establish the factor of one half; that factor emerges from the motion relationship or the corresponding work integral.

For a constant available driving power, F = P/v means that the possible driving force decreases as speed rises. Meanwhile, air resistance usually increases with speed. Their intersection gives a possible steady top speed under those conditions. This model assumes the quoted power reaches the driven motion and that the road gradient is included in the resistance balance. The same vehicle can have a different top speed uphill, in a headwind or with different transmission losses. Balanced forces do not imply zero power; they imply no further kinetic-energy increase.

Check that the horizontal axis is displacement, not time. A force–time area is impulse and describes momentum change, whereas a force–displacement area is work and describes energy transfer. For a graph of the pulling force, the area gives that force’s work; subtract other resistive work before identifying net kinetic-energy gain. Divide complicated graphs into recognisable shapes or narrow strips, retaining negative regions if the force opposes displacement. The gradient of this graph is not the work: its units are force per length, which can instead describe a spring stiffness.

Choose one reference and use it consistently for every state. If a ball falls from a shelf to the floor, selecting the floor as zero makes the final gravitational term vanish; selecting the shelf as zero makes the final term negative. Both give the same potential-energy decrease and therefore the same motion when other assumptions match. The approximation mgΔh assumes g is effectively constant over the height change. For orbital distances, use the radial gravitational-potential expression with its usual zero at infinity instead of extending a surface formula too far.

Identify what counts as useful before writing the fraction. For a motor lifting a load, the useful output might be the gain in gravitational potential energy, while the input is electrical energy. Quoting the total energy supplied to the motor as the useful output would conceal heating losses. Average and instantaneous powers should not be mixed if the operation changes with time. Efficiencies of successive stages multiply as fractions, because each stage receives only the previous stage’s useful output. A stated efficiency of 80% means 0.80 in that multiplication, not 80.