Efficiency and mechanical energy

A bicycle has stopped, and its brake pads have become hot. Has what previously enabled it to move and do work disappeared? We will follow energy from raised and moving objects to heating, and assess what fraction of the input achieves our goal.

Raise a load and release it: as it descends, it can lift another object through a rope. Its new position has therefore given it the ability to do work. Interaction energy that depends on the relative positions of objects is called potential energy. For the load, it belongs to the object–Earth system. A compressed spring can also do work as it expands; its potential energy is associated with deformation. Visible rest does not mean an absence of energy.

From which height should we measure the load's energy: the floor or the tabletop? Choose a zero level. Measure height h vertically from it to the centre of mass, a point describing mass distribution; for a uniform rectangular block, it is at the centre. Moving the zero changes the numerical potential energy. But for the same lift, the height change, required work and energy change remain unchanged.

Slowly lifting a load through height h requires overcoming its approximately constant gravitational force mg. The lifting work is mgh, so relative to the chosen zero, Eп = mgh for constant g near Earth's surface. Doubling mass or height doubles the energy; at h = 0 it is zero by our choice of reference. kg·N/kg·m gives joules. This formula describes gravity, not a spring; energy change depends on height change, not on the length of a detour.

A moving ball can shift an object, and a moving cart can compress a spring while slowing down. Energy associated with motion is kinetic energy Eк. It tells us how much work can be obtained during braking under suitable conditions. Motion is measured in a chosen reference frame: a ball on a bus seat is at rest relative to the bus but moves relative to the road. Kinetic energy therefore depends on the frame and is zero when the object is at rest in that frame.

For translational motion in school mechanics, Eк = mv²/2. Why does speed have a stronger effect than mass? Under the same braking force, a faster object takes longer to lose speed and covers distance at a greater average speed throughout; braking work grows for both reasons. The formula quantifies the result: doubling mass doubles energy, while doubling speed quadruples it. Mass in kg and speed in m/s give energy in J.

Release a load from rest. As it falls, height and Eп decrease while speed and Eк increase. If only gravity acts and air resistance is negligible, the loss of one exactly equals the gain of the other: Eп + Eк = const. Near the ground, chosen as the zero level, almost all the initial mechanical energy has become kinetic. This relates two states without calculating position every second.

Now include air or brakes. Eп + Eк may decrease, but energy does not disappear. Some becomes internal energy of the objects and surroundings: energy of their particles' motion and interactions. Hot brakes make this visible. 'Loss' refers to our goal: the energy is less available for the original mechanical work. A complete balance includes the mechanism and surroundings as well as the moving load.

A machine has a task, such as lifting a load onto a shelf. Call the work serving that goal Aпол. Call all work supplied to the machine Aзатр: it includes the useful result, heating and other transformations. Both are measured in joules. First define the goal and boundaries of the process. Deformation may be a loss for a lift but a useful result for a press; no physical process is inherently labelled 'useful'.

What fraction of the input served the goal? Divide useful work by input work: η = Aпол/Aзатр for Aзатр > 0. This is efficiency, a dimensionless fraction; multiply by 100% to express it as a percentage. At fixed input, more useful work raises efficiency; reducing losses brings it closer to 100%. The ideal limit is 100%; a real machine with losses falls below it. Above 100% means this balance has omitted an energy source or defined the work incorrectly.

A machine lifts a 2 kg load through 3 m using 100 J. Take g = 10 Н/кг, with equal initial and final speeds. Find the efficiency.

The useful result is the load's increase in potential energy: Aпол = mgh = 2·10·3 = 60 Дж.

Efficiency: η = 60/100 = 0,6 = 60%.

The remaining 40 J has not disappeared: it went into other transformations, such as heating the machine.

Efficiency: 60%.

Potential energy is associated with interaction and position, kinetic energy with motion: Eп = mgh and Eк = mv²/2 under the stated conditions.

Without resistance, mechanical energy is conserved during a fall; friction converts some of it into internal energy.

Efficiency compares the useful result with total input. 'Losses' mean a different route of energy transformation, not energy disappearing.

In grade 8, we will explain internal energy and thermal processes through particle motion and interactions.