Energy, work and power

Work done by a constant force Paper 4

A force does work when its point of application moves. For a constant force $F$ whose point of application moves a distance $d$ in a straight line,

$$\text{work done}=Fd\cos \theta$$

where $\theta$ is the angle between the force and the displacement. Only the component of the force along the displacement, $F\cos \theta$, does work; the component at right angles does none. Work is a scalar, measured in joules: $1\text{J}$ is the work done when a force of $1\text{N}$ moves its point of application $1\text{m}$ along its own direction. (The scalar product is not needed for this syllabus.)

• If $\theta =0$, the work is simply $Fd$.
• If $\theta ={90}^{°}$, no work is done. The normal reaction on a particle sliding along a surface does no work, and neither does the weight of a particle moving horizontally.
• If the force opposes the motion ($\theta ={180}^{°}$), the work done by it is $-Fd$. We usually say instead that $Fd$ of work is done against it, as with friction and air resistance.

Work done against gravity in raising a particle through a height $h$ is $mgh$, whatever path it takes, because only the vertical part of the displacement is along the weight.

Only the component $F\cos \theta$ acts along the displacement $d$, so the work done by $F$ is $Fd\cos \theta$.

Kinetic and gravitational potential energy Paper 4

Energy is the capacity to do work, and is also measured in joules.

• Kinetic energy (KE) is the energy of motion: $\text{KE}=\frac{1}{2}m{v}^2$. It depends on speed, not direction, and doubling the speed multiplies it by four.
• Gravitational potential energy (PE) is the energy of position: $\text{PE}=mgh$, where $h$ is the height above a level you choose as zero. Only changes in PE matter, so choose the zero level wherever is convenient, often the lowest point of the motion.

Where these formulae come from: if a constant resultant force $F$ accelerates a mass $m$ from rest through a distance $s$, then $v^2=2as$ and $F=ma$, so the work done is $Fs=ma\times \frac{v^2}{2a}=\frac{1}{2}m{v}^2$. The work has become kinetic energy. Similarly, lifting a mass at steady speed through $h$ needs work $mgh$, which is stored as potential energy.

The work-energy principle and conservation of energy Paper 4

The two key results:

Work-energy principle. The change in the total mechanical energy (KE $+$ PE) of a system equals the work done on it by the external forces other than weight. Written out for a typical problem:

$$\text{work done by driving force}-\text{work done against resistance}=\text{gain in KE}+\text{gain in PE}$$

Conservation of energy. If no force other than weight does any work (the surfaces are smooth, there is no air resistance, and any normal reaction is perpendicular to the motion), the total mechanical energy is constant: $\text{KE}+\text{PE}$ at the start $=$ $\text{KE}+\text{PE}$ at the end.

Energy methods are the tool of choice when the motion is not in a straight line, for example a child on a smooth curved slide or a ball on a curved track. Newton's second law would need the forces at every point, but energy needs only the start and the end: only overall energy changes need to be considered. They are also quick for straight-line problems where you want a speed or a distance rather than a time.

Always write the energy equation in words first, then put in numbers. State which forces do work, and check that each appears exactly once.

Power Paper 4

Power is the rate at which a force does work, measured in watts: $1\text{W}=1{\text{J s}}^{-1}$, and $1\text{kW}=1000\text{W}$.

• Average power over a time: $\text{power}=\frac{\text{work done}}{\text{time taken}}$.
• Power of a force in the direction of motion: in a short time $\delta t$ the force does work $F\delta s$, so the rate of working is $F\frac{\delta s}{\delta t}$, which tends to

$$P=Fv$$

For a vehicle, $P$ is the power of the engine and $F$ is the driving force. At a fixed power, the driving force is $F=\frac{P}{v}$: large at low speed, small at high speed. That is why a car accelerates hard from low speeds and then less and less.

A vehicle reaches its maximum speed when it can no longer accelerate: the driving force just balances the resistances (and the weight component on a hill). On level ground, $\frac{P}{v_{\text{max}}}=\text{resistance}$. Always convert kilowatts to watts first.

Cars on hills: instantaneous acceleration Paper 4

On a hill inclined at $\alpha$, the weight has a component $mg\sin \alpha$ along the road, against the car going uphill and with it downhill. Newton's second law along the road, for a car going up:

$$\frac{P}{v}-R-mg\sin \alpha =ma$$

where $R$ is the resistance. Because $\frac{P}{v}$ changes with speed, this gives the instantaneous acceleration at that speed; the acceleration is not constant, so do not use the constant acceleration formulae for this motion. The maximum speed up the hill is where $a=0$. Going down, $mg\sin \alpha$ changes sign.

Questions often state the slope as $\sin \alpha =\frac{1}{20}$ or similar, so $mg\sin \alpha$ is easy to find without the angle.