Q.A cyclist comes to a skidding stop in $10\ \text{m}$. During this process, the force on the cycle due to the road is $200\ \text{N}$ and is directly opposed to the motion.

Concept: Work-Energy Theorem — work done by a force equals the change in kinetic energy, and the sign depends on whether the force aids or opposes motion.

(a) The road exerts a friction force of $200\ \text{N}$ opposite to the displacement of $10\ \text{m}$. Work done by a constant force is $W = F d \cos\theta$, where $\theta$ is the angle between force and displacement. Here $\theta = 180^\circ$, so $\cos 180^\circ = -1$.

$W_\text{road on cycle} = (200\ \text{N})(10\ \text{m})(-1) = -2000\ \text{J}$ …

Why the Work–Energy Theorem is the natural lens

Work is defined as $W = \vec{F} \cdot \vec{s} = F s \cos\theta$, where $\theta$ is the angle between the force and the displacement of the point the force acts on. When a force opposes motion, $\theta = 180^\circ$ and $\cos\theta = -1$, so the work is negative. That negative work is exactly what removes kinetic energy from the cyclist — the road's friction force does negative work on the cycle, bringing it to a stop.

But the question also asks about the work by the cycle on the road. That's where Newton's third law comes in: the force on the road is equal in magnitude and opposite in direction to the force on the cycle. The crucial subtlety is what displacement to use for this second calculation — and it is not the cycle's own displacement.

Step-by-step solution

1. Identify the force and displacement for part (a)

The road exerts a friction force of $200\ \text{N}$ directly opposite to the motion. The cycle moves $10\ \text{m}$ in the direction of motion. So the angle between the force (backward) and displacement (forward) is $180^\circ$.

2. Compute work done by the road on the cycle

The negative sign tells us the road is removing energy from the cycle — exactly what friction does.

3. For part (b), apply Newton's third law

The force on the road due to the cycle is equal in magnitude and opposite in direction to the force on the cycle due to the road: $200\ \text{N}$, forward.

4. Identify the correct displacement to use for "work done on the road"

Work is always the force times the displacement of the material point on which the force acts. The force from the cycle acts on the road's surface — the actual asphalt/ground material at the contact patch. That material point of the road does not move: the road is rigid and fixed to the Earth. It is the contact location that shifts along the road as the tyre slides, not any single piece of the road's material. …

Concept: Work Done by a Force Opposing Motion + Newton's Third Law

Step 1: Work done by the road on the cycle.

The road's friction force ($200\,\text{N}$) is directly opposite the $10\,\text{m}$ displacement, so $\theta = 180^\circ$:

Step 2: Identify the reaction force on the road (Newton's third law).

The cycle exerts a $200\,\text{N}$ force on the road, equal and opposite to the force the road exerts on the cycle.

Step 3: Identify the correct displacement for "work done on the road." …