Work and energy

Work is done on a body only when a force succeeds in moving it, and only the part of the force that acts along the direction of motion counts.

where: $F$ is the applied force (newton, N), $s$ is the displacement of the body (metre, m), and $\theta$ is the angle between the force direction and the displacement direction. The SI unit of work is the joule (J), defined by $1\ \text{J} = 1\ \text{N} \times 1\ \text{m}$ — one joule of work is done when a force of one newton moves its point of application through one metre in its own direction. Two special angles decide almost every ICSE question on this formula. When $\theta = 0°$, force and displacement point the same way, $\cos 0° = 1$, and $W = Fs$ — the largest work possible for that force and distance. When $\theta = 90°$, force and displacement are perpendicular, $\cos 90° = 0$, and $W = 0$ even though the object is genuinely moving and a real force is genuinely acting. A student carrying a schoolbag horizontally does no work against gravity, because the weight of the bag acts straight down while the walking motion is horizontal — the two directions are at right angles. A common special case worth separating out is work done against gravity. To raise a body of mass $m$ through a height $h$ at (roughly) constant speed, the applied force must equal the weight $mg$, acting straight up, over a displacement $h$ straight up — so $\theta = 0°$ and:

This quantity is exactly the gravitational potential energy the body gains, which is why "work done in lifting" and "PE gained" are numerically the same story told two ways. Work and energy share the joule, but several other named units appear on the ICSE syllabus purely as conversions — no new physics, just different-sized measuring cups for the same quantity:

• $1\ \text{J} = 10^{7}\ \text{erg}$ — the CGS unit of work/energy.
• $1\ \text{kWh} = 3.6 \times 10^{6}\ \text{J}$ — the commercial "unit" on an electricity bill.
• $1\ \text{cal} = 4.186\ \text{J}$ — the calorie, used for heat energy.
• $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ — the electronvolt, used for atomic-scale energy.
• $1\ \text{hp} = 746\ \text{W}$ — a power unit (not energy), included here because it is often confused with the others.

Energy is the capacity of a body to do work. The two ideas are locked together by a single accounting rule: whenever work is done on a body, it gains that much energy; whenever a body does work, it loses that much energy. This "work done equals energy transferred" rule is the thread that later connects kinetic energy, potential energy and the principle of conservation of energy, all of which build directly on the formula above.

A boy pulls a trolley with a force of 50 N along a horizontal floor through a distance of 12 m, with the force applied at 60° to the floor. Calculate the work done, and compare it with the work a crane does lifting a 250 kg load through 6 m.

Step 1. List the knowns for the trolley: $F = 50$ N, $s = 12$ m, $\theta = 60°$.

Step 2. Apply $W = Fs\cos\theta$: $W = 50 \times 12 \times \cos 60° = 50 \times 12 \times 0.5 = 300\ \text{J}$. Only the horizontal component of the pull, $50\cos 60° = 25$ N, actually contributes to moving the trolley; the vertical component does no work since it is perpendicular to the floor.

Step 3. Now the crane: $m = 250$ kg, $h = 6$ m, $g = 9.8\ \text{m/s}^2$. Apply $W = mgh$: $W = 250 \times 9.8 \times 6 = 2450 \times 6 = 14{,}700\ \text{J}$.

Step 4. Compare. The crane does $14{,}700\ \text{J}$, roughly 49 times the boy's 300 J — reasonable, since lifting 250 kg against full gravity dwarfs dragging a light trolley at an angle. The boy does 300 J of work on the trolley; the crane does 14,700 J (14.7 kJ) of work against gravity, which is stored as gravitational potential energy in the raised load.

• Thinking work is done whenever force is applied and effort is felt — pushing an immovable wall does zero work, however hard you push, because there is no displacement.
• Forgetting that only the component of force along the displacement counts — at $\theta = 90°$, $W = 0$ even with a large force and large displacement, as with carrying a bag horizontally.
• Mixing units — plugging ergs, calories or kWh straight into $W = Fs\cos\theta$ without first converting to joules gives a numerically wrong answer.
• Confusing "work done by gravity" (positive as a body falls) with "work done against gravity" (positive as a body is raised) — always check which direction the question describes.

Ask the tutor: want a mixed set of $W = Fs\cos\theta$ and $W = mgh$ problems, or help seeing why lifting a bag straight up counts as work but carrying it flat across a room does not?