Work and Power

Definition. Work done = force applied × distance moved IN THE DIRECTION of the force.

$W = F \times d$

• $W$ in joules (J)
• $F$ in newtons (N)
• $d$ in metres (m)

1 J = 1 N m. One joule is the work done when a 1 N force moves an object 1 m in the direction of the force.

Force must be parallel to motion. If a force acts at an angle, only the COMPONENT of the force along the direction of motion does work. A horizontal force lifting nothing (no vertical motion) does no work against gravity; the weight of a horizontally-moving trolley does no work because weight is perpendicular to the motion.

Spec 4.12: work done = energy transferred. Whenever a force does work on an object, energy is transferred TO or FROM the object. The kind of energy transferred depends on the situation:

• Pushing a box along a floor: work done by the push transfers to KE (if accelerating) and/or to thermal energy of the floor/box via friction.
• Lifting a box vertically at constant speed: work done by the lift force transfers to the GPE store of the box.
• Stretching a spring: work done by the pulling force transfers to elastic PE of the spring.

Sign convention. Work done BY a force in the direction of motion is POSITIVE; work done AGAINST the motion (friction, drag) is NEGATIVE — it removes energy.

Definition. An object lifted vertically gains gravitational potential energy (GPE) because work is done against the force of gravity.

$\Delta GPE = m \times g \times \Delta h$

• $m$ in kg
• $g$ ≈ 9.8 N/kg (often rounded to 10 N/kg in 4PH1 questions — check the question)
• $\Delta h$ = VERTICAL height change in m

Reference level. GPE is always measured RELATIVE to a chosen zero — usually ground level. Only CHANGES in GPE matter physically.

Worked sanity check. Lifting a 1 kg textbook by 1 m: $\Delta GPE = 1 \times 10 \times 1 = 10$ J. Lifting it through 2 m doubles the GPE.

Slant paths. If you push a trolley up a RAMP of length $L$ at angle $\theta$, the height gained is $h = L \sin\theta$. Use this $h$ in $mgh$. The work done by the push along the ramp uses $L$ (the slant distance).

Air resistance / friction free? If the object is RAISED at constant speed, the work done by the applied force equals the GAIN in GPE — assuming no friction. If friction is significant, some applied work goes to thermal energy.

Definition. Energy stored by a moving object.

$KE = \dfrac{1}{2} m v^2$

• $m$ in kg
• $v$ in m/s (the SPEED — direction unimportant for KE)

The square is crucial. Doubling speed QUADRUPLES the KE. Trebling speed makes 9× the KE. This is why high-speed crashes are disproportionately destructive — and why fuel-economy plummets at motorway speeds.

Worked sanity check. A 1 kg ball at 1 m/s: $KE = 0.5 \times 1 \times 1^2 = 0.5$ J. At 10 m/s: $KE = 0.5 \times 1 \times 100 = 50$ J (×100, even though speed only ×10).

Practical implications:

• Braking distance. The KE of a car must be removed by friction at the wheels. Doubling speed → 4× KE → braking distance roughly quadruples (assuming constant braking force).
• Crash energy. A modern car at 60 km/h has roughly 4× the crash energy of one at 30 km/h. Crumple zones must absorb that energy via deformation (work done against internal forces).

Conservation of energy says the total energy of a closed system is constant. The most common 4PH1 application links GPE, KE and work done.

Falling object (no air resistance). A ball dropped from height $h$:

$\text{GPE at top} = \text{KE at bottom}$

$m g h = \tfrac{1}{2} m v^2$

Cancel $m$ → $v = \sqrt{2gh}$. Mass-independent: a heavy ball and a feather hit the ground at the same speed (in vacuum). Galileo's classic Leaning Tower of Pisa story.

Pendulum. Energy oscillates between GPE (at the extremes of swing) and KE (at the bottom). With no friction, the bob always returns to the same height. In practice, air resistance drains energy to thermal store and the swing decays.

Roller-coaster. At the highest point, mostly GPE. At the lowest point, mostly KE. Some energy is wasted to thermal store via friction → cars don't quite reach the same height as their starting point.

Work-energy theorem. When a force does work on an object, the work done equals the change in the object's energy. If the only energy change is kinetic: $W = \Delta KE$.

Stopping a vehicle. The braking force does NEGATIVE work, removing KE. Setting $|W| = KE$ gives $F \times d = \tfrac{1}{2}mv^2$. Hence braking distance $d \propto v^2$ for constant force.

Definition. Power is the RATE at which work is done (= rate at which energy is transferred).

$P = \dfrac{W}{t} = \dfrac{\Delta E}{t}$

• $P$ in watts (W)
• $W$ or $\Delta E$ in joules (J)
• $t$ in seconds (s)

1 W = 1 J/s. One watt is the power needed to do 1 joule of work in 1 second.

Common power values.

• Domestic light bulb: 5-100 W (LED 5-15 W; halogen 35-100 W).
• Kettle: 2-3 kW.
• Domestic electric heater: 1-3 kW.
• Sports car: ~150 kW.
• National-grid power station: 500-2000 MW.

Climbing stairs. A 60 kg person climbing 3 m of stairs in 4 s does $W = mgh = 60 \times 10 \times 3 = 1800$ J. Power $= 1800/4 = 450$ W ≈ ½ horsepower.

Power from force × velocity (advanced). If a constant force $F$ moves at constant speed $v$, power output = $F \times v$. (This isn't on the spec but examiners sometimes use it implicitly: e.g. car traveling at 30 m/s against drag of 500 N requires 15 kW of engine power.)

Don't confuse power with energy. A 100 W lamp draws 100 J every second. Over 10 s, that's 1000 J. Power tells you the RATE; multiplying by time gives the total energy.