Variable forces and powerAQA A-Level Further Maths: Revision notes

When the force $F$ depends on position $x$ (and acts along the line of motion), the work done in moving from $x=a$ to $x=b$ is $W=\int_a^bF\,dx,$ the area under the force–displacement graph. If $F$ is constant this reduces to $W=F(b-a)$. Example: $F=3x^2+2$ from $x=0$ to $x=4$: $W=[x^3+2x]_0^4=72$ J. From $x=1$ to $x=3$: $W=(27+6)-(1+2)=30$ J.

The net work done on a particle equals its change in kinetic energy, even when the force varies: $\int_a^bF\,dx=\frac12mv^2-\frac12mu^2.$ This avoids solving a differential equation. Example: a 4 kg particle passes $O$ at $2\text{ m s}^{-1}$ under $F=20-4x$. Work from $x=0$ to $x=3$ is $[20x-2x^2]_0^3=42$ J, so $\frac12(4)v^2=8+42=50$ and $v=5\text{ m s}^{-1}$. The speed is greatest where the force changes sign, here at $x=5$: for $x<5$ the force speeds the particle up and for $x>5$ it slows it down. The work to $x=5$ is $50$ J, so $\frac12(4)v^2=58$ and $v=\sqrt{29}=5.39\text{ m s}^{-1}$.

Power is the rate of doing work: $P=\frac{dW}{dt}.$ The unit is the watt (W), equal to 1 J s⁻¹; 1 kW $=1000$ W. For a force $F$ acting in the direction of motion at speed $v$ (AS: motion in a straight line along the force) $P=Fv.$ Always convert kilowatts to watts before substituting. Example: an engine of 30 kW at $20\text{ m s}^{-1}$ provides a driving force $F=\frac{30\,000}{20}=1500$ N.

For a vehicle on a horizontal road, the driving force from the engine is $F=\frac{P}{v}$ and Newton's second law gives $\frac{P}{v}-R=ma.$ As the speed increases, $F=\frac{P}{v}$ decreases, so the acceleration falls. At the maximum speed the acceleration is zero, so $F=R$ and $v_{\max}=\frac{P}{R}$. Example (car, 1200 kg, 30 kW, $R=750$ N): at $20\text{ m s}^{-1}$, $1500-750=1200a$ gives $a=0.625\text{ m s}^{-2}$; the maximum speed is $\frac{30\,000}{750}=40\text{ m s}^{-1}$.

When the power changes, the speed cannot change instantly, so the new driving force is the new power divided by the speed at that instant. Then use $F-R=ma$. Example: a train (200 000 kg, $R=8000$ N) is at steady $50\text{ m s}^{-1}$ on 400 kW. The power drops to 300 kW: the force becomes $\frac{300\,000}{50}=6000$ N, so $6000-8000=200\,000a$ and the deceleration is $0.01\text{ m s}^{-2}$. The train slows until $F=R$ again, at $v=\frac{300\,000}{8000}=37.5\text{ m s}^{-1}$.