A-LevelMathematicsMechanics9709/42 · Oct/Nov 2023 · Q35 Marks
A block of mass 10kg is at rest on a rough plane inclined at an angle of 30° to the horizontal. A force of 120N is applied to the block at an angle of 20° above a line of greatest slope (see diagram). There is a force resisting the motion of the block and 200J of work is done against this force when the block has moved a distance of 5 m up the plane from rest. [Figure 3]
- Find the speed of the block when it has moved a distance of 5 m up the plane from rest.[5]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [B1] Work done by 120N force = 120×5cos 20[=563.81557....]
- [B1] (PE change =)10g×5sin30[=250] For PE change.
- [M1] Attempt at work energy equation 4 relevant terms; dimensionally correct; allow sign errors; allow sin/cos mix in relevant resolved terms.
- [A1] 120×5cos20-10g×5sin30-200=½×10×v² [563.815...-250-200 = 5v²]
- [A1] Speed = 4.77 ms⁻¹ awrt 4.77.
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →