Work and Energy

Definition. The work done by a constant force $\mathbf{F}$ over a displacement $\mathbf{d}$ is

$W = \mathbf{F} \cdot \mathbf{d} = F d \cos\theta,$

where $\theta$ is the angle between force and displacement. Units: joules (J), where $1\,\text{J} = 1\,\text{N m}$.

Special cases.

• $\theta = 0^{\circ}$: force along displacement, $W = Fd$ (maximum positive work).
• $\theta = 90^{\circ}$: force perpendicular to displacement, $W = 0$ (e.g. normal reaction does no work on a body moving along a surface).
• $\theta = 180^{\circ}$: force opposite to displacement, $W = -Fd$ (e.g. friction does negative work on a sliding body).

Work done by a system of forces. Total work = sum of works done by each force. Equivalently, work done by the resultant force:

$W_{\text{net}} = \sum_{i} W_{i} = \mathbf{F}_{\text{net}} \cdot \mathbf{d}.$

Worked example. A $30$ N force pushes a box $8$ m horizontally; friction is $19.6$ N opposing motion.

• Work by applied force: $W_{\text{app}} = 30 \cdot 8 = 240$ J.
• Work by friction: $W_{\text{fric}} = -19.6 \cdot 8 = -156.8$ J.
• Net work: $240 - 156.8 = 83.2$ J.

(Vertical forces — weight and normal reaction — do no work since the box moves horizontally.)

Kinetic energy of a particle of mass $m$ at speed $v$:

$\text{KE} = \tfrac{1}{2}mv^{2}.$

Always non-negative. Depends on speed (scalar), not velocity (vector).

Gravitational potential energy at height $h$ above a chosen reference:

$\text{PE} = mgh.$

Only changes in PE are physically meaningful. The reference level is your choice — pick the most convenient one (typically the lowest point of the motion).

Total mechanical energy. $E = \text{KE} + \text{PE}$. For an isolated system with only conservative forces (gravity), $E$ is constant.

Why the work-energy theorem. Newton's second law $F = ma$ for a particle moving in 1D. Multiply by $v\,dt = ds$:

$F\,ds = m a\,ds = m\,dv \cdot \dfrac{ds}{dt} = m v\,dv.$

Integrating: $\int F\,ds = \int m v\,dv$, i.e. $W = \tfrac{1}{2}m v^{2} - \tfrac{1}{2}m u^{2}$.

Application. Whenever there is no easy suvat path (variable force, change of direction, curved path), use the work-energy theorem.

Smooth (frictionless) motion. The total mechanical energy is conserved:

$\tfrac{1}{2}m u^{2} + m g h_{1} = \tfrac{1}{2}m v^{2} + m g h_{2}.$

The mass cancels for a single particle — useful for finding $v$ from height changes.

With friction (or other non-conservative forces). The work done against the resistive forces is the missing energy:

$\underbrace{\tfrac{1}{2}m u^{2} + m g h_{1}}_{\text{initial mech. energy}} = \underbrace{\tfrac{1}{2}m v^{2} + m g h_{2}}_{\text{final mech. energy}} + W_{\text{against friction}}.$

Worked example — smooth slope. Particle of mass $0.5$ kg, length $4$ m, $30^{\circ}$ slope, from rest.

• $h = 4\sin 30^{\circ} = 2$ m.
• $mgh = \tfrac{1}{2}mv^{2}$ gives $v = \sqrt{2 \cdot 9.8 \cdot 2} \approx 6.26$ m/s.

Worked example — rough slope. Box of mass $3$ kg, length $10$ m, $25^{\circ}$, $\mu = 0.2$, from rest.

• $h = 10\sin 25^{\circ} \approx 4.226$ m.
• $N = 3g\cos 25^{\circ} \approx 26.64$ N.
• $F_{\text{fric}} = 0.2 N \approx 5.33$ N.
• $W_{\text{fric}} = 5.33 \cdot 10 \approx 53.3$ J.
• Energy equation: $3 \cdot 9.8 \cdot 4.226 = \tfrac{1}{2}(3)v^{2} + 53.3$, i.e. $124.2 = 1.5 v^{2} + 53.3$.
• $v^{2} = 47.3$, $v \approx 6.88$ m/s.

Why use energy methods? Often faster than suvat or Newton's second law on curved paths, multi-step motions, or when only speeds (not times) are wanted.

Power is the rate of doing work:

$P = \dfrac{dW}{dt} = F v$

for a force $F$ acting in the direction of motion at speed $v$. Units: watts (W); $1$ kW = $1000$ W; $1$ hp $\approx 746$ W (not used in IAL).

Vehicle moving under constant engine power. The driving (tractive) force the engine exerts on the road through the wheels is

$F_{\text{drive}} = \dfrac{P}{v}.$

So driving force decreases as the vehicle speeds up. This is the key insight for max-speed problems.

Maximum speed. Acceleration is zero at $v_{\max}$:

• Newton's 2nd law: $F_{\text{drive}} - F_{\text{resist}} = ma = 0$.
• So $F_{\text{drive}} = F_{\text{resist}}$ at top speed.
• Substitute: $P/v_{\max} = F_{\text{resist}}$, giving $v_{\max} = P/F_{\text{resist}}$.

Worked example — level road. Car of mass $1200$ kg, $P = 24$ kW, resistance $400$ N.

• $v_{\max} = 24000/400 = 60$ m/s.
• At $v = 30$ m/s: $F_{\text{drive}} = 24000/30 = 800$ N. Net force $= 800 - 400 = 400$ N. Acceleration $a = 400/1200 = 1/3$ m/s$^{2}$.

Worked example — incline (uphill). Van of mass $1800$ kg, $\sin\theta = 1/14$, non-grav resistance $750$ N, $P = 30$ kW.

• At $v_{\max}$: $F_{\text{drive}} = 750 + 1800g\sin\theta = 750 + 1260 = 2010$ N.
• $v_{\max} = 30000/2010 \approx 14.9$ m/s.

Driving downhill. Gravity component now helps the vehicle: $F_{\text{drive}} + mg\sin\theta = F_{\text{resist}}$ at terminal speed.

Standard question structures on WME02/01.

Type 1: Slope with friction (energy method). A particle slides $d$ metres down (or is projected up) a rough slope. Find (a) work done against friction, (b) speed at the end. Energy conservation with friction loss is the cleanest route.

Type 2: Vehicle on level road. Constant power $P$. Resistance $R$. Find (a) max speed, (b) acceleration at a given speed. The two parts use the same machinery: $F = P/v$, then $ma = F - R$.

Type 3: Vehicle on a hill. Same as Type 2 but with an inclination $\theta$ — add or subtract $mg\sin\theta$ to the resistance.

Type 4: Combined (work-energy theorem on a vehicle). A car ascends a hill from one speed to another over a given distance, with constant power and constant resistance. Use the energy equation: $W_{\text{net}} = \Delta\text{KE}$, where $W_{\text{net}}$ includes work done by the engine, by gravity, and against friction. This is the hardest standard variant — typically the highest-mark part of the energy question.

Marking pattern. Force diagrams unlock M-marks. Energy bookkeeping must be careful — sign errors on $W_{\text{against gravity}}$ or $W_{\text{against friction}}$ are the #1 mark losses.