A block of mass 4kg is pulled along a rough horizontal road by a constant force of magnitude 25 N acting at an angle of 36° above the horizontal. The block moves in a straight line passing through two points A and B on the road, where AB = 120m. The coefficient of friction between the block and the road is 0.4.
- (a)Find the work done against friction in moving the block from A to B.[3]
- (b)The speed of the block at A is 7 ms⁻¹. Use an energy method to find the speed of the block at B.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.
- [1] Attempt to resolve vertically to form an equation for the normal contact force for the block; correct number of dimensionally correct terms, condone sign errors, and sin/cos mix on the 25 N force term. Use of R = 4g scores no marks in this part.
- [1] Use of Work done = 0.4R×120 ONLY so M0 if subtracting/adding the Work done by the pulling force to this value e.g. a final answer of 1212 (or 1210) coming from 120×(25cos36-0.4×(40-25sin36)) scores M1 M0 A0.
- [1] CWO 1210 or final AWRT 1215 or 1214 (usually coming from using R = 25.3). For reference: 1214.6576...
- [1] SOI, by KE terms on opposite sides of an equation with the same sign so must imply a change in KE (e.g. 2v² on one side and 98 on the other).
- [1] Work Done by pulling force = ±25cos36×120 [=±2427.050...]
- [1] Attempt at work-energy equation using their work done against friction from 3(a) with four dimensionally correct terms but allow sign errors and sin/cos mix on the pulling force term. Dependent on both M1 marks in 3(a) unless the total work done is recovered in this part e.g. a candidate might, in 3(a), find the total work done by the block (which if correct is 1212.39...) rather than the required work done against friction. Therefore, an answer of 1212.39... 3(a) would score M1 M0 A0 but then seeing 2v²-98=1212.39... (or even 1210 but only if coming from 1212.39...) followed by 25.6 would score full marks in this part. M0 if using 25×120 for the pulling force.
- [1] vB = 25.6m s⁻¹ AWRT 25.6. (For reference: 25.59680...). (Special case for using N2L: [1] This is for getting a=2.53. (For reference: a = 2.525819...). [1] AWRT 25.6. (For reference: 25.59680...).)
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