| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Moderate -0.3 This is a straightforward application of the power-work-energy relationship with standard mechanics formulas. Part (i) requires simple multiplication (Power × time), while part (ii) involves resolving forces on an incline and using P=Fv, but follows a standard textbook pattern with no novel problem-solving required. Slightly easier than average due to the routine nature of both parts. |
| Spec | 6.02a Work done: concept and definition6.02l Power and velocity: P = Fv |
Total: 8 marks
**Part (i):**
$WD = P \times t = 75 \times 60$ **M1**
$= 4500$ kJ or $4500000$ J oe **A1** Allow 4500000 for A1 but not 4500.
**Part (ii):**
$v = 24 \text{ ms}^{-1}$ so $F = P/v$ **M1*** Find resistive force (could be found in (i))
$= 3125$ **A1** 3125 seen
N2 with $P/v$ (used here or later) and component of weight **M1***
$\dfrac{75000}{v} - 0.05 \times 4000g - 3125 = 0$ **A1** Correct equation
Solve for $v$ **M1dep***
Hence $v = 14.6 \text{ ms}^{-1}$ **A1** Answer in range [14.6, 14.7]
Total: **8 marks**10 The engine of a lorry of mass 4000 kg works at a constant rate of 75 kW . Resistance to motion is modelled by a constant resistive force. On a horizontal road the lorry travels at a constant speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the work done by the engine in travelling for 1 minute on the horizontal road.\\
(ii) The lorry travels at a constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a slope of angle $\sin ^ { - 1 } 0.05$ to the horizontal. Find the value of $v$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2017 Q10 [8]}}