A block of mass 5 kg is placed on a plane inclined at 30° to the horizontal. The coefficient of friction between the block and the plane is μ.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

1. [4] (a) R = 5g cos 30 [= 25√3]
2. [4] (a) 40 - 5g sin 30 - F > 0
3. [4] (a) State that the net force up the plane is positive, 3 terms
4. [4] (a) F = µ × 5g cos 30
5. [4] (a) For using F = µR with R as a component of 5g to obtain an equality/inequality in µ only with 3 terms
6. [4] (a) µ < 1/√3
7. [4] (a) AG
8. [4] (a) Alternative scheme for question 6(a)
9. [4] (a) R = 5g cos 30 [= 25√3]
10. [4] (a) 40 - 5g sin 30 – F = 5a
11. [4] (a) Acceleration a > 0
12. [4] (a) F = µ × 5g cos 30
13. [4] (a) For using F = µR with R as a component of 5g to obtain an equality in µ and a
14. [4] (a) µ < 1/√3
15. [4] (a) AG. From µ = (40 - 5g sin 30 - 5a) / (5g cos 30) with a > 0
16. [4] (b) Attempt to resolve forces parallel to or perpendicular to the inclined plane, 3 relevant terms in either direction
17. [4] (b) R = 5g cos 30 + 40 sin 30 [= 20 + 25√3 = 63.3]
18. [4] (b) F = 40 cos 30 – 5g sin 30 [= 20√3 – 25 = 9.64]
19. [4] (b) µ ≥ 0.152
20. [4] (b) AG. Using F < µR
21. [4] (b) Alternative method for question 6(b)
22. [4] (b) Attempt to resolve forces horizontally or vertically with 3 relevant terms in either direction
23. [4] (b) 40 = R sin 30 + F cos 30 [40 = ½R + √3/2F]
24. [4] (b) 5g = R cos 30 – F sin 30 [5g = √3/2R – ½F]
25. [4] (b) µ ≥ 0.152
26. [4] (b) AG. Solve for R and F and use F < µR