5.1 Energy conservation

For a constant force, work done is W=Fs when the displacement s is along the force. More generally W=Fs cosθ for angle θ.

Work is energy transferred, so choose the force and displacement pair and use the component of force along the displacement.

A 40 N force moving a box 3.0 m in its direction transfers 120 J; a perpendicular force transfers zero work.

A force can act without doing work if there is no displacement or if it is perpendicular to the motion.

The principle of conservation of energy says total energy is constant: energy is transferred between stores, not created or destroyed.

Define the system boundary and identify where energy enters, leaves or changes store. Include useful and dissipated pathways in a complete account.

A falling object transfers gravitational potential energy mainly to kinetic energy, then to thermal energy through drag.

“Wasted” energy has not vanished; it has usually spread into less useful thermal stores.

Efficiency is η=useful output energy ÷ total input energy, often expressed as a percentage.

Use the same quantity type on top and bottom: useful energy with input energy, or useful power with input power. The remainder is not destroyed.

A motor receiving 500 J and delivering 350 J of useful mechanical energy has efficiency 0.70, or 70%.

Efficiency cannot exceed 100% for a passive system, and useful output is not necessarily the whole output.

Rearrange η=E_useful/E_input to find an unknown energy or power, then calculate the non-useful share as input minus useful output.

Convert percentages to decimals before substituting and label the direction of the calculation. Check that the answer is physically plausible.

At 80% efficiency, a 2.0 kW useful output requires 2.5 kW input; 0.5 kW is dissipated.

Do not multiply by 100 twice, and do not use dissipated energy as the denominator unless the question defines it as input.

Power is the rate of energy transfer: P=W/t=∆E/t, measured in watts, where 1 W=1 J s⁻¹.

Use the energy transferred over the relevant time interval; average power need not equal instantaneous power when the rate changes.

A device transferring 900 J in 30 s has average power 30 W.

Power is not the same as total work: two machines can do the same work in different times and therefore have different powers.

For a process with work W completed in time t, average power is P=W/t. Rearranging gives W=Pt or t=W/P.

Keep time in seconds for watts and identify whether W is useful work or total input work before substituting.

A 1.2 kW lift doing 36 kJ of work takes 30 s if its power is constant.

A larger power does not mean more total work unless the operating time is also considered.

When a force component F acts along an object’s velocity v, power transferred is P=Fv; more generally use the parallel component.

Use the resultant or driving force specified by the problem and keep the direction/sign clear. This relation follows from work per time because v=s/t.

A 400 N driving force moving a vehicle at 15 m s⁻¹ supplies 6.0 kW of mechanical power.

P=Fv is not P=ma; it relates force to energy-transfer rate and requires the velocity component along the force.