5. Work, energy and power

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.

For a near-Earth height change ∆h, work done against gravity is ∆E_P=mg∆h. Raising an object transfers energy to its gravitational potential store.

Use the vertical height change, not the length of a sloping path, when g is uniform. A downward move reverses the sign of the store change.

Lifting 2.0 kg through 3.0 m at g=9.8 m s⁻² increases gravitational potential energy by 58.8 J.

The energy change depends on height difference, not route length; friction would add a separate thermal transfer.

In a uniform field, the change in gravitational potential energy is ∆E_P=mg∆h, with g treated as constant over the height interval.

Define the reference level and keep the sign of ∆h consistent. Only differences in potential energy affect energy conservation calculations.

A 0.50 kg mass lowered 4.0 m has ∆E_P=−19.6 J relative to its starting level; that energy can become kinetic or thermal.

Zero potential at the floor is a choice, not a physical claim that the object has no energy anywhere else.

The constant-acceleration equations give links between u, v, a, s and t; kinetic energy is E_K=½mv² and changes when the net work changes speed.

Check that acceleration is constant before selecting an SUVAT equation, then use the signed velocity and the correct mass in the energy calculation.

An object accelerating from rest at 2.0 m s⁻² for 3.0 s reaches 6.0 m s⁻¹, so its kinetic energy is 18m joules for mass m.

SUVAT is not valid for arbitrary changing acceleration, and kinetic energy depends on speed squared, not velocity sign.

The kinetic-energy store of a mass m moving at speed v is E_K=½mv². It is a scalar and is never negative.

Use speed magnitude and consistent units; doubling speed quadruples kinetic energy, while doubling mass doubles it.

A 4.0 kg trolley moving at 3.0 m s⁻¹ has kinetic energy 18 J.

Kinetic energy does not carry the direction sign of momentum, and stopping does not destroy it—it transfers it to other stores.