| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Moderate -0.3 This is a straightforward energy conservation problem with friction on an inclined plane. Students must equate initial KE + loss in PE to final KE + work done against friction, then solve for the friction force. It requires standard mechanics techniques (energy methods, resolving on slopes) with no novel insight, making it slightly easier than average for A-level mechanics. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
PE lost $= 0.4g \times 3\sin 20°\ [4.104]$
Initial KE $= \tfrac{1}{2}\times 0.4\times 0.5^2\ [0.05]$
Final KE $= \tfrac{1}{2}\times 0.4\times 2.5^2\ [1.25]$
Difference $=$ Work done by friction
$2.9045 = 3F$
Therefore $F = \mathbf{0.968}$ N
- M1: $mgh$ attempted, with trig
- M1: Both KEs attempted
- M1: Work/Energy principle used, no extra/missing terms
- M1: Use $\Delta E = F \times s$
- A1: Answer, a.r.t. 0.968 [SC: Energy not used: answer B2]
**[5]**8 A rough plane is inclined at $20 ^ { \circ }$ to the horizontal. A particle of mass 0.4 kg is projected down the plane, along a line of greatest slope, at $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. After it has travelled 3 m down the plane its speed is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. By considering the change in energy, find the magnitude of the frictional force, assumed constant.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q8 [5]}}