Work, energy and powerAQA A-Level Physics: Revision notes

Work done is the energy transferred when a force moves its point of application. For a constant force at an angle to the displacement:

$W = Fs\cos\theta$

where $F$ is the force (N), $s$ the displacement (m) and $\theta$ the angle between the force and the direction of motion. Work is measured in joules (J); 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force.

$F\cos\theta$ is the component of the force along the motion; the perpendicular component does no work. Example: 40 N at 30° moving a suitcase 25 m does $40\times25\times\cos30^\circ=870$ J. If $\theta=90^\circ$ the work is zero.

When the force changes, the work done is the area under a force-displacement graph. Split the area into triangles and rectangles.

Example: a force rises uniformly from 0 to 80 N over 5.0 m, then stays at 80 N for 3.0 m. Work = $\tfrac12\times5.0\times80+80\times3.0=200+240=440$ J. If friction is negligible, this equals the gain in kinetic energy: $\tfrac12mv^2=440$ J, so for a 20 kg crate $v=6.6$ m s⁻¹.

The same idea applies to the extension of a spring, where the force-extension graph is a straight line through the origin.

Power is the rate of doing work, or the rate of energy transfer:

$P=\dfrac{W}{t}$

Measured in watts (W), where 1 W = 1 J s⁻¹. If a force $F$ moves an object at constant velocity $v$ in the direction of the force, the work done per second is $Fv$:

$P=Fv$

Example: a car at constant 25 m s⁻¹ with resistive force 600 N needs a driving force of 600 N, so $P=600\times25=15$ kW. At constant speed the driving force equals the resistive force; if the speed rises the force needed for a given power falls.

Efficiency compares useful output with total input:

$\text{efficiency}=\dfrac{\text{useful output power}}{\text{input power}}$

It can also be written with energy instead of power. It is a ratio with no unit, or a percentage when multiplied by 100. It can never exceed 100%, and the energy not usefully transferred is dissipated, usually as thermal energy.

Example: a car delivering 15 kW to the wheels from fuel supplying energy at 60 kW has efficiency $15\div60=0.25$ (25%). For a cyclist climbing at 536 W useful power with 24% efficiency, the chemical energy supplied per second is $536\div0.24=2.2$ kW.

A cyclist and bicycle of total mass 85 kg climb a 6.0° hill at constant 5.0 m s⁻¹ against 20 N of resistive force.

1. Weight component down the slope: $mg\sin\theta=85\times9.81\times\sin6.0^\circ=87$ N.
2. Constant speed means resultant force is zero, so the driving force is $87+20=107$ N.
3. Useful power: $P=Fv=107\times5.0=5.4\times10^2$ W.

For a motor delivering only 250 W the greatest speed is $v=P/F=250\div107=2.3$ m s⁻¹.