Scientific conception of work

Two conditions, both required. Physics says work is done on a body only when a force acts on it and the body is displaced as a result of that force. Neither condition alone is enough — a taut force with zero displacement does no work, and so, oddly, does a "cause" that is not actually responsible for the motion. This is the single rule the whole topic hangs on.

A scalar, measured in joules. Work has no direction of its own — it is a scalar quantity, positive, negative, or zero, never "north" or "up." Its SI unit is the joule (J), built directly from force and distance:

For the simplest case — a constant force acting along the direction of motion — the amount of work is just force times displacement:

where: $W$ is the work done (joule, J), $F$ is the magnitude of the constant force (newton, N), and $s$ is the distance moved by the object in the direction of that force (metre, m). One joule is the work done when a force of one newton moves an object through one metre along the force's own direction.

Reading off the sign. When force and displacement point the same way, $W$ is positive (gravity accelerating a falling stone). When they point opposite ways, $W$ is negative (friction resisting a sliding book, or gravity acting on a ball thrown upward). When the force is perpendicular to the displacement, or the displacement is zero, $W = 0$ — no cosine formula is needed at this stage to see either of those two special cases.

At Class 9 level, work stays a purely definitional idea rather than a formula you derive from deeper laws — so instead of a derivation, here is the reasoning behind why the two-condition test and the joule are set up the way they are, tying the qualitative rule to the one quantitative formula this stage allows: a force acting exactly along the line of motion does work equal to $F \times s$, and this single case already explains every result used in this chapter.

Step 1. Start from what "doing work" should mean physically: transferring energy into or out of a moving object. A force that never moves anything cannot be handing over any energy — however hard it pushes, nothing changes about the object's motion or position, so bookkeeping "zero work" is the only consistent choice.

Step 2. Now take the case where the force does act along the direction of travel. If a constant force $F$ pushes an object a distance $s$ in its own direction, common sense says doubling the force should double the work, and doubling the distance should also double it. The simplest quantity with that property is the product $F \times s$, which is exactly the definition $W = F \times s$ used above.

Step 3. Check the unit against the definition: $F$ is in newtons and $s$ is in metres, so $W$ comes out in newton-metres, which is renamed the joule for convenience. This confirms $1\ \text{J} = 1\ \text{N}\,\cdot\,\text{m}$ is not a separate fact to memorise — it falls straight out of $W = F \times s$.

Step 4. Extend the reasoning to force not along the direction of motion, but only for the two special angles tested at this stage. When the force is exactly perpendicular to the displacement, none of that force acts along the path travelled, so by the same "no motion along the force" logic as Step 1, the work done by that force is zero — this is why a coolie's upward supporting force does zero work on a load carried horizontally. When the force and the displacement point in exactly opposite directions, the force is resisting the motion rather than driving it, and the work done by that force is counted as negative, subtracting energy from the moving object rather than adding it.

A porter carries a suitcase of weight 150 N on his head and walks 20 m along a level platform at constant speed — reusing the earlier reasoning that his vertical supporting force does zero work, since it is perpendicular to the horizontal displacement. Now suppose the same suitcase is instead dragged along the ground by a rope, pulled with a constant horizontal force of 50 N over the same 20 m, with the rope kept parallel to the ground. Find the work done by the pulling force.

Step 1. Identify the two conditions: a force of $F = 50\ \text{N}$ acts on the suitcase, and it is displaced $s = 20\ \text{m}$ because of that pull. Both conditions for work are satisfied, unlike the earlier vertical-carry case.

Step 2. Check the direction: the rope's pull is horizontal, and the suitcase's displacement is also horizontal, in the same line. This is the simple case where $W = F \times s$ applies directly, with no perpendicular or opposing component to worry about.

Step 3. Substitute the values: $W = F \times s = 50 \times 20 = 1000\ \text{J}$.

Step 4. Conclusion: the pulling force does $W = 1000\ \text{J}$ of positive work on the suitcase — very different from the porter's carrying force, which did zero work on the identical 20 m journey, purely because of the angle between force and displacement, not the size of the force.

• Treating "trying hard" or "getting tired" as proof that work is being done — physics work needs actual displacement caused by the force, not effort.
• Forgetting that work can be negative, not just zero — a force directly opposing the motion (friction, or gravity on a rising ball) removes energy and counts as negative work.
• Applying $W = F \times s$ when the force is perpendicular to the displacement — that formula is only for force along the direction of motion; a perpendicular force does zero work regardless of its size.
• Thinking work is a vector because force and displacement are vectors — work itself is a scalar, and its sign (plus or minus) only tells you whether energy is added or removed, not a direction.

Ask the tutor: Describe any situation with a force and a motion, and ask the AI tutor to check with you whether the work done is positive, negative, or zero using the two-condition rule.