Work

Two conditions, both required. A force must act on the object, and the object must actually move because of that force. Miss either one and, in the physics sense, zero work is done — no matter how much effort was spent.

The formula. For a constant force $F$ acting on an object that moves a distance $s$ in the same direction as the force:

where: $W$ is the work done, in joules (J); $F$ is the applied force, in newtons (N); and $s$ is the displacement of the object in the direction of the force, in metres (m). This chapter uses only the case where force and motion point the same way — the angled case, $W = Fs\cos\theta$, belongs to Class 10.

The unit. One joule is the work done when a force of 1 newton moves an object through 1 metre in its own direction:

Work is a scalar — magnitude only, no direction — and it shares its unit with energy for a good reason: doing work is precisely how energy is handed from one object or store to another.

Sign, without the trigonometry. If a force acts in the direction the object is already moving, it does positive work on the object and hands it energy — a push that speeds up a trolley. If a force such as friction opposes the motion, it does negative work and removes energy, slowing the object down. If there is no displacement at all, the work done is exactly zero. At this level these signs are recognised qualitatively; only the direct case $W = Fs$ is ever calculated as a number.

There is no algebraic derivation needed at this level — the formula $W = F \times s$ is a definition, and the reasoning behind it is physical rather than mathematical. Force alone tells you how hard something is being pushed or pulled, but it says nothing about the effect that push has on the object. Displacement alone tells you how far something has moved, but says nothing about what caused it. Work combines the two, because it is meant to measure how much a force actually achieves — and a force that produces no movement achieves nothing on the object, however large it is. This is why the formula multiplies rather than adds the two quantities, and why doubling either one doubles the work: pushing twice as hard over the same distance transfers twice the energy, and pushing the same box twice as far transfers twice the energy too. It is also why work and energy carry the same unit, the joule — work is defined precisely as the amount of energy transferred to (or taken from) an object by a force, so tracking the work done on something is the same as tracking the energy it gains or loses.

Question: A boy pushes a box with a constant horizontal force of 25 N, and the box slides 4 m across the floor in the direction of the push. Later, his friend pushes the same box with a force of 40 N against a wall, but the box does not move at all. Find the work done in each case.

Step 1. Identify what is given for the first push: $F = 25\ \text{N}$, $s = 4\ \text{m}$, and the force and displacement are in the same direction, so the direct formula applies.

Step 2. Apply $W = F \times s$ for the first push: $W = 25 \times 4 = 100\ \text{J}$.

Step 3. Check the second push against the two conditions for work. A force of $40\ \text{N}$ does act on the box, satisfying the first condition — but the box does not move, so $s = 0$, and the second condition fails.

Step 4. Apply the formula anyway, with $s = 0$: $W = 40 \times 0 = 0\ \text{J}$.

The boy does 100 J of work on the box by sliding it 4 m; his friend does 0 J of work on the box, however hard he pushes, because the box never moves.

• Assuming that feeling tired or straining means work is being done — in physics, work needs actual displacement of the object, not just muscular effort.
• Forgetting that a force acting with no resulting movement (like pushing a wall) always gives zero work, regardless of how large the force is.
• Mixing up the newton (unit of force) with the joule (unit of work and energy) — they measure different quantities and are never interchangeable.
• Trying to use $W = Fs\cos\theta$ for angled forces at this level — only the same-direction case $W = Fs$ is required in this chapter; the angled case is Class 10 syllabus.
• Overlooking that carrying a schoolbag at a steady height while walking involves an upward supporting force but horizontal displacement, so the supporting force itself does no work in the direction of motion.

Ask the tutor: Still unsure why holding something heavy counts as zero work? Ask the AI tutor for more everyday examples and quick practice problems on $W = F \times s$.