Work done by a constant force

Push a trolley across a level floor and something measurable happens even though the trolley itself has no memory of the push: energy has been transferred from your muscles into its motion. Physics needs a precise number for "how much" of that transfer occurred, and for the simplest case — a constant force acting along a straight-line path — the answer turns out to be a single multiplication. This article gives that formula, W=F×sW = F \times s, defines its unit the joule, and works through the sign rules that decide whether a force adds energy to a moving object or removes it. This is the calculation engine of the CBSE Class 9 "Work and Energy" chapter (Unit III, Motion, Force and Work). Every numerical you meet in this topic — an applied push, a resisting friction, a rising or falling weight — reduces to this one formula, used carefully with the right sign. Only forces acting exactly along or exactly against the direction of motion are considered at this level; the general case with an angle between force and displacement belongs to Class 11.

The core insight

The work done by a constant force acting along the line of motion is W=F×sW = F \times s, positive when force and displacement point the same way and negative when they oppose, and its SI unit is the joule.

The core idea

1. The formula. When a constant force FF acts on an object and the object moves a distance ss in the direction of that force, the work done by the force is defined as:

W=F×sW = F \times s

where:

  • WW = work done, in joules (J)
  • FF = magnitude of the constant force, in newtons (N)
  • ss = magnitude of the displacement caused by (or occurring in the direction of) that force, in metres (m)

2. The unit. The SI unit of work is the joule (J). One joule is the work done when a force of 1 newton moves an object through 1 metre in the direction of the force:

1 J=1 N×1 m=1 N ⋅ m1\ \text{J} = 1\ \text{N} \times 1\ \text{m} = 1\ \text{N}\,\cdot\,\text{m}

3. The sign rule. Because a force can either help or hinder motion, work carries a sign that must be reasoned out, not just calculated:

  • Force and displacement in the same direction → work is positive, W=+FsW = +Fs. Energy is being added to the object (an applied push speeding a trolley up).
  • Force and displacement in opposite directions → work is negative, W=−FsW = -Fs. Energy is being removed (friction slowing a sliding crate, or gravity acting while an object rises).
  • No displacement at all, or force perpendicular to the displacement, gives W=0W = 0 — no energy is transferred by that force, however large it is.

4. Work is a scalar. Work has a magnitude and a sign but no direction of its own — it is a scalar quantity, unlike force and displacement, which are vectors. The sign of WW tells you whether energy flows into or out of the object, not which way it "points."

Figure
A trolley pushed by an arrow labelled force F moving a distance s in the same direction, next to a crate opposed by a friction arrow pointing backward, illustrating positive work when force and displacement align and negative work when they oppose
Same direction gives positive work, opposite direction gives negative work Applied force, same direction — W = +F s F displacement s Friction, opposite direction — W = -f s motion s f
Derivation AWork done by a constant applied force, W = F times s

The goal is to establish, from the plain-language definition of work as "force applied over a distance," the exact formula W=F×sW = F \times s for a constant force acting along the direction of motion, together with the direction rule that turns the bare product into a signed quantity — the single tool used throughout this chapter for every numerical.

Step 1. Start from the definition: work is done whenever a force produces a displacement. If no displacement occurs — however hard you push against a locked door — no work is done, because there is nothing for the force to act "through." Work therefore must depend on both the force FF and the displacement ss it causes, not on either alone.

Step 2. Fix the simplest case: force and displacement lie along the same straight line, in the same sense. Experiments (and the definition adopted for this level) show the work done grows in direct proportion to the force applied and in direct proportion to the distance moved — doubling either one doubles the work. The only combination with this property is the product, so W=F×sW = F \times s.

Step 3. Fix the unit by the definition of the joule: substituting F=1 NF = 1\ \text{N} and s=1 ms = 1\ \text{m} gives W=1 N×1 m=1 JW = 1\ \text{N} \times 1\ \text{m} = 1\ \text{J}, which is exactly how the joule is defined. This confirms the formula is dimensionally and numerically consistent with the SI unit of work.

Step 4. Extend to the opposing case by physical reasoning, not new algebra: if the force acts against the direction of motion (like friction on a sliding object), it is removing kinetic energy rather than adding it, so the same magnitude F×sF \times s is assigned a negative sign, W=−FsW = -Fs. The formula W=F×sW = F \times s together with this direction rule covers every constant-force, straight-line case tested at this level. Check: for the applied-force case F=25 NF = 25\ \text{N}, s=6 ms = 6\ \text{m} gives W=150 JW = 150\ \text{J} (positive, same direction); for a resisting force f=60 Nf = 60\ \text{N}, s=10 ms = 10\ \text{m} gives W=−600 JW = -600\ \text{J} (negative, opposing direction) — both confirmed in the worked example below.

Worked example

A horizontal force of 25 N pushes a trolley 6 m along a straight, level floor, in the direction of the push. Separately, a crate is dragged 10 m along a rough floor while a constant frictional force of 60 N opposes its motion. Find the work done by the applied force on the trolley, and the work done by friction on the crate.

Step 1. Trolley — identify the quantities: F=25 NF = 25\ \text{N}, s=6 ms = 6\ \text{m}, force and displacement are in the same direction, so the sign will be positive.

Step 2. Trolley — apply the formula: W=F×s=25 N×6 m=150 N ⋅ m=150 JW = F \times s = 25\ \text{N} \times 6\ \text{m} = 150\ \text{N}\,\cdot\,\text{m} = 150\ \text{J}. Since the force pushes the trolley the way it moves, W=+150 JW = +150\ \text{J}.

Step 3. Crate — identify the quantities: f=60 Nf = 60\ \text{N}, s=10 ms = 10\ \text{m}, friction acts opposite to the direction of motion.

Step 4. Crate — apply the formula for magnitude, then assign the sign: Wfriction=f×s=60 N×10 m=600 JW_{friction} = f \times s = 60\ \text{N} \times 10\ \text{m} = 600\ \text{J}; because friction opposes the crate's displacement, Wfriction=−600 JW_{friction} = -600\ \text{J}. This means 600 J of the crate's energy is drained away as heat at the rubbing surfaces over that 10 m.

Final answers: work by the applied force on the trolley =+150 J= +150\ \text{J}; work by friction on the crate =−600 J= -600\ \text{J}.

Common traps

  • Using the total path length instead of the displacement caused by (or aligned with) the specific force being considered — if a force acts over only part of the path, only that part counts.
  • Dropping the sign: reporting friction's work as +600 J+600\ \text{J} because the multiplication gives a positive number, instead of reasoning that friction opposes motion and so removes energy, giving −600 J-600\ \text{J}.
  • Skipping unit conversion — plugging in centimetres or grams-based forces instead of metres and newtons, which silently corrupts the joule answer; always convert to SI first.
  • Assuming a bigger force automatically means more work — work depends equally on displacement, so a huge force producing zero or negligible displacement does zero or negligible work.

Ask the tutor: Share a problem with several forces acting on one object (an applied push plus friction, say) and ask the AI tutor to help you set up W=F×sW = F \times s with the correct sign for each force in turn.

If I push a box and it does not move, have I really done zero work, even though I got tired?

Yes, in the physics sense of work, even though your muscles used chemical energy and produced fatigue. The formula W=F×sW = F \times s needs an actual displacement ss of the object the force acts on; with s=0s = 0, W=0W = 0 regardless of how hard you pushed or for how long. Your tiredness comes from continuous muscle contraction and internal energy use, which is a biological cost, not mechanical work transferred to the box. This is one of the sharpest differences between the everyday meaning of "work" and its precise physics definition.

Why is the work done by friction always negative, but the work done by an applied push often positive?

The sign depends only on the angle between the force and the displacement, not on which force it is. An applied push that moves an object in its own direction has force and displacement aligned, so the work is positive — energy flows into the object. Friction, by its nature, always acts opposite to the relative motion it resists, so its displacement is opposite to its own force direction, making its work negative — energy flows out of the object as heat. It is not that friction is "special," only that its direction is always opposing by definition.

Do I need to worry about the angle between force and displacement in Class 9?

No, not at this level. Class 9 restricts every problem to forces acting exactly along the line of motion, either helping it (positive work) or opposing it (negative work), or exactly perpendicular to it (zero work). The full formula that handles a force at any angle, W=Fscos⁡θW = Fs\cos\theta, is introduced only in Class 11; you will meet the sign rule you are learning now as the special case θ=0\theta = 0 degrees or θ=180\theta = 180 degrees of that later formula.

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