Work done and energyAQA A-Level Further Maths: Revision notes

The work done by a constant force acting in the direction of motion is $W=Fd,$ where $F$ is the force in newtons and $d$ the distance moved in metres, in the direction of the force. The unit is the joule (J). Work is a scalar. If the force opposes the motion (a resistance, friction), it does negative work on the particle; we talk about the work done against the resistance, $Fd$, which is a positive amount of energy lost. The net work done on a particle is the sum of the work done by all the forces, and is the work done by the resultant force. Example: a crate is pulled 12 m by a 70 N force against a 30 N resistance. Work done by the pull $=840$ J, work against resistance $=360$ J, net work $=480$ J.

The kinetic energy of a particle of mass $m$ moving at speed $v$ is $KE=\frac12mv^2.$ The work–energy principle states that the net work done on a particle equals its change in kinetic energy: $\text{net work done}=\frac12mv^2-\frac12mu^2.$ This is often quicker than using forces and $F=ma$, because it needs no direction information and no acceleration. Example: the crate above starts from rest, so $480=\frac12(20)v^2$ and $v=\sqrt{48}=6.93\text{ m s}^{-1}$.

The gravitational potential energy (GPE) of a particle of mass $m$ at height $h$ above a chosen reference level is $GPE=mgh.$ Only the change matters: when a particle rises a vertical height $h$ it gains $mgh$, and when it falls it loses $mgh$. On a slope of length $d$ inclined at $\theta$ to the horizontal, the vertical height is $h=d\sin\theta$, so the change in GPE is $mgd\sin\theta$. Always use the vertical height, not the distance along the slope.

If only gravity does work (no resistance, no driving force), mechanical energy is conserved: $\frac12mu^2+mgh_1=\frac12mv^2+mgh_2.$ Example: a ball thrown up at $14\text{ m s}^{-1}$ has $39.2$ J of kinetic energy, so it rises $\frac{39.2}{0.4\times9.8}=10$ m. At 6 m, $\frac12(0.4)v^2=39.2-23.52$ and $v=8.85\text{ m s}^{-1}$. When other forces do work, use the general energy equation: $\text{initial KE}+\text{initial GPE}+\text{work by driving forces}=\text{final KE}+\text{final GPE}+\text{work against resistances}.$ Example: the skier sliding 80 m down a $20^\circ$ slope against a 50 N resistance loses $18\,770$ J of GPE, does $4000$ J of work against resistance, and so gains $14\,770$ J of kinetic energy.

On a rough surface the friction force is $F=\mu R$, with $R$ found by resolving perpendicular to the surface (on a plane inclined at $\theta$, $R=mg\cos\theta$). Friction always opposes motion, so it does negative work, and the work done against it is $\mu R\,d$ over a path of length $d$. The whole path length counts. Example: a particle of mass 0.5 kg projected up a $30^\circ$ plane ($\mu=0.3$) at $8\text{ m s}^{-1}$: $F=0.3\times0.5(9.8)\cos30^\circ=1.27$ N, and $16=2.45d+1.27d$ gives $d=4.30$ m. For the whole round trip the net change in GPE is zero, so only friction changes the energy: $\frac12(0.5)v^2=16-1.27\times2\times4.30$ and $v=4.50\text{ m s}^{-1}$, less than the starting speed.