A block of mass 8kg slides down a rough plane inclined at 30° to the horizontal, starting from rest. The coefficient of friction between the block and the plane is μ. The block accelerates uniformly down the plane at 2.4 ms⁻².

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [B1] Correct force diagram with 3 forces in the correct directions. No labels required on the 3 forces and ignore wrong labels. Arrows needed. Allow either or both components of weight if fully labelled. Allow sin/cos mix. If forces are not connected to the block, then the line of action of each force must go through the block.
2. [B1] R=8gcos30 [= 40√3=69.282...] Resolving perpendicular to the plane.
3. [M1*] Resolving parallel to the plane and attempt to apply Newton's second law. 3 terms. Allow sign errors, sin/cos mix. Allow g missing, otherwise dimensionally correct. ([8g sin 30-F = 8×2.4 ⇒F = 20.8])
4. [DM1] Use of F = µR to get an equation in μ only. Allow g missing in either or both of F and R. Allow sign errors, consistent sin/cos mix. R must be a single component of a force. Allow the 3 masses to be cancelled. ([8g sin 30-8gcos30=8×2.4 40-40√3μ=19.2])
5. [A1] μ = 0.3[0...] [May first see 20.8/(40√3) or 20.8/69.282...] Allow exact value (13-√3)/75 or 104√3/600 oe.
6. [B1] [v² = 2×2.4×3 ⇒ greatest speed =] 3.79 ms⁻¹ = 6√10/5 3.79473... (3.8 without a more accurate value seen gets B0 and should be annotated SF).