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

The kinetic energy of a body of mass $m$ moving at speed $v$ is $\frac12mv^2$. The gravitational potential energy of a body at height $h$ above a chosen reference level is $mgh$. Both are measured in joules (J), with $g=9.8$ m s$^{-2}$. Kinetic energy depends on speed only, never on direction, so it can never be negative. Example: a $1200$ kg car moving at $30$ m s$^{-1}$ has $\frac12(1200)(30^2)=540\,000$ J. A change in kinetic energy is $\frac12m(v^2-u^2)$, not $\frac12m(v-u)^2$.

The work done by a constant force $F$ moving its point of application a distance $d$ in the direction of the force is $Fd$. If the force makes an angle $\theta$ with the direction of motion, work $=Fd\cos\theta$. Work is a scalar, measured in joules. Work done against a force (gravity, friction, resistance) is positive when the motion is opposed to the force. Up a slope of angle $\alpha$ through a distance $d$, the work done against gravity is $mg\,d\sin\alpha$, because the vertical rise is $d\sin\alpha$. Friction does work $F_rd$ against the motion, where $F_r=\mu R$ for limiting friction and $R=mg\cos\alpha$ on a slope.

Power is the rate of doing work, $P=\frac{\text{work}}{t}$, measured in watts (W): one joule per second. For a vehicle with driving force $F$ at speed $v$: $P=Fv.$ Use this to find $F$ at a given speed, then apply Newton's second law: $F-R=ma$ on a level road, or $F-R-mg\sin\alpha=ma$ up a slope. At the maximum or constant speed, $a=0$, so $F=R$ (or $R+mg\sin\alpha$ up a slope). Convert kW to W before using the equation.

The work done by all the forces on a body equals its change in kinetic energy. In the form that is easiest to use: $\text{work done by driving forces}-\text{work done against resistances}=\text{change in kinetic energy}+\text{change in potential energy}.$ If a particle slides up a slope against gravity and friction, the work-energy equation is $\frac12mu^2-\frac12mv^2=mgd\sin\alpha+Fd$. For the return journey down, gravity does positive work and friction negative work: $\frac12mv^2=mgd\sin\alpha-Fd$.

On an incline: (1) resolve perpendicular to the plane to find $R=mg\cos\alpha$; (2) find friction $F=\mu R$; (3) write a work-energy equation using the distance along the slope; (4) solve. Example: $0.4$ kg launched at $12$ m s$^{-1}$ up a $30^\circ$ slope with $\mu=0.25$: $28.8=(1.96+0.849)d$, so $d=10.3$ m. For vehicles: find $F=\frac Pv$, then use the work-energy principle for a distance, or Newton's second law for acceleration. Give answers with the unit (J, W, m s$^{-1}$) and to 3 significant figures.