A cyclist is riding a bicycle along a straight horizontal road AB of length 50m. The cyclist starts from rest at A and reaches a speed of 6ms⁻¹ at B. The cyclist produces a constant driving force of magnitude 100N. There is a resistance force, and the work done against the resistance force from A to B is 3560J.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. [B1] Work done by cyclist = 50 × 100 (= 5000 J)
2. [M1] Work energy equation. Three terms (Work Done by driving force - Work Done against resistance = Change in Kinetic Energy). Allow sign errors. Dimensionally correct. Their 5000 – 3560 = ½ mx6²
3. [A1] mass = 80 kg
4. [SC M1] Alternative Method: Acceleration considered as a constant. From use of v=6, u=0, s=50. Must be using correct suvat formulae. 6² = 0 + 2a × 50 [⇒ a = 0.36]. For equation involving mass using N2L with three terms (Driving force - Resistance / Distance = mass × acceleration). Allow sign errors in N2L. 100 – 3560/50 = m × their 0.36
5. [SC A1] mass = 80 kg