Work and power with forces at an angleAQA A-Level Further Maths: Revision notes

If a constant force $F$ makes an angle $\theta$ with the direction of motion, only the component along the motion does work: $W=Fd\cos\theta,$ where $d$ is the distance moved. If $\theta=90^\circ$ the work done is zero (normal reactions do no work). If $\theta>90^\circ$ the work done is negative, for example a resistance, $\theta=180^\circ$. Example: a sledge is pulled 30 m by a 40 N rope at $25^\circ$ above the horizontal: $W=40\times30\times\cos25^\circ=1090$ J.

Combine $W=Fd\cos\theta$ with the work–energy principle: net work done $=$ change in kinetic energy. Add the work of each force separately, including friction and the resolved component of the pull. Example: the sledge has a 28 N horizontal resistance. Net work $=1087.6-28\times30=247.6$ J, so $\frac12(15)v^2=247.6$ and $v=5.75\text{ m s}^{-1}$. The weight and the normal reaction do no work on horizontal ground, because both are at $90^\circ$ to the motion.

On a smooth ramp inclined at $\alpha$, the work done against gravity when a box moves a distance $d$ up the ramp is $mgd\sin\alpha$ (the weight component down the slope times $d$); the gravitational potential energy gain is the same. A rope at angle $\beta$ to the ramp does work $Fd\cos\beta$. Example: a 20 kg box, 6 m up a $20^\circ$ ramp, pulled by 120 N at $30^\circ$ to the ramp: rope work $=120\times6\cos30^\circ=624$ J, GPE gain $=20(9.8)(6\sin20^\circ)=402$ J, so $\frac12(20)v^2=221$ and $v=4.70\text{ m s}^{-1}$.

The rate at which a force does work is $P=Fv\cos\theta,$ where $\theta$ is the angle between the force and the velocity; this is $Fv$ when the force is along the motion. In vehicle problems the driving force acts along the road, but the weight has a component along the slope, so you must resolve to find the driving force needed. Example: a 40 N rope at $25^\circ$ to the horizontal pulling a sledge at $5\text{ m s}^{-1}$ develops $40\times5\times\cos25^\circ=181$ W.

For a vehicle moving up a slope with $\sin\theta$ known, the driving force $F=\frac{P}{v}$ along the road must overcome the resistance $R$ and the weight component $mg\sin\theta$: $\frac{P}{v}-R-mg\sin\theta=ma.$ At steady speed $a=0$ and the greatest speed is $v=\frac{P}{R+mg\sin\theta}$. Example: lorry, 5000 kg, $\sin\alpha=\frac{1}{25}$, $R=1500$ N, $P=80$ kW. $F=1500+1960=3460$ N and $v=\frac{80\,000}{3460}=23.1\text{ m s}^{-1}$. At $15\text{ m s}^{-1}$, $\frac{80\,000}{15}-3460=5000a$ gives $a=0.375\text{ m s}^{-2}$. At constant speed the work done by the engine over a distance $s$ is $(R+mg\sin\theta)s$, which equals the gain in potential energy plus the work done against resistance.