Work Done by Frictional Force
Example 5.3
A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion.
Given Values
Force (F) = 200 N
Distance (d) = 10 m
Angle (θ) = 180°
Calculated Work
W = F × d × cosθ
W = 200 × 10 × (-1)
W = -2000 J
(a) Work done by the road on the cycle
W = Fd cosθ = 200 × 10 × cos(180°) = 200 × 10 × (-1) = -2000 J
This negative work brings the cycle to a halt in accordance with the Work-Energy theorem.
(b) Work done by the cycle on the road
From Newton's Third Law, an equal and opposite force (200 N) acts on the road due to the cycle.
However, the road undergoes no displacement (d = 0), so:
W = F × d = 200 × 0 = 0 J
Key Concept
While the forces between two bodies are always equal and opposite (Newton's Third Law), the work done on each body is not necessarily equal and opposite.
Work depends on both force and displacement, and the displacement of each body may be different.
