Work, energy and powerEdexcel A-Level Further Maths: Revision notes

The work done by a constant force $F$ moving its point of application a distance $d$ in the direction of the force is $Fd$ (in joules). If the force makes an angle $\theta$ with the motion, the work done is $Fd\cos\theta$. Kinetic energy is $\frac12mv^2$ and gravitational potential energy is $mgh$ above a chosen level. For a $0.2$ kg stone thrown up at $15$ m s$^{-1}$, $\text{KE}=\frac12(0.2)(15)^2=22.5$ J.

The work-energy principle states that the total work done on a particle equals its change in kinetic energy. More usefully for problems: $\text{work done by driving forces}-\text{work done against resistances}=\Delta\text{KE}+\Delta\text{PE}.$ Friction on a rough surface is $F=\mu R$ when the particle is moving, with $R=mg\cos\alpha$ on a slope. For a $4$ kg particle projected up a plane at $30^\circ$ with $\mu=0.3$ and speed $8$ m s$^{-1}$: $F=0.3(4)(9.8)\cos30^\circ=10.18$ N, so $128=(19.6+10.18)d$ and $d=4.30$ m.

If the only forces doing work are gravity (and a smooth surface does none), total mechanical energy is conserved: $\text{KE}+\text{PE}$ stays constant. For the stone at $5$ m: $\frac12(0.2)v^2+0.2(9.8)(5)=22.5$, giving $v=11.3$ m s$^{-1}$. Whenever friction, air resistance or a driving force is present, include their work instead of using conservation alone.

Power is the rate of doing work: $P=\frac{\text{work}}{\text{time}}$, measured in watts (W). For a force $F$ acting on an object moving with speed $v$ in the direction of the force, $P=Fv.$ For a vehicle, $F$ is the driving force from the engine. Use Newton's second law along the direction of motion: $F-R=ma$, where $R$ is the resistance. At maximum speed the acceleration is zero, so $F=R$. A car of $24$ kW against $600$ N has maximum speed $\frac{24\,000}{600}=40$ m s$^{-1}$.

When the resistance varies with speed, for example $R=100v$, the driving force $F=\frac{P}{v}$ also varies, so acceleration changes. Use Newton's second law at each speed: $\frac{P}{v}-R-mg\sin\theta=ma$ going up a slope. At constant speed, set $a=0$. For $P=40\,000$ and $R=100v$ up a slope with $mg\sin\theta=2450$: $\frac{40\,000}{v}=100v+2450$, a quadratic in $v$ giving $v=11.2$ m s$^{-1}$. Down a slope with the engine off, the weight component acts forwards: $2450=100v$ gives $v=24.5$ m s$^{-1}$.