| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Maximum speed on incline vs horizontal |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring application of work-energy principles and power equations. Part (a) involves basic energy conservation with given values (finding distance from work done against resistance, then applying energy equation), while part (b) is a standard power-on-slope calculation. All techniques are routine for Further Maths mechanics with no novel problem-solving required. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) Work-energy principle: \(16 \times d = 1440(000)\) | M1 | Used, \(F \times d = E\) |
| \(d = 90(000)\) (km (m)) | A1 | cao, oe |
| (ii) Using expression for PE or KE: | M1 | |
| At end, \(KE = \frac{1}{2}(80)(10)^2 = 4000\) | A1 | \(= 4000\) J \(= 4\) kJ |
| Answer | Marks |
|---|---|
| \end{cases}\] | A1 |
| Work-energy principle (See notes): All terms included, oe FT their KE and PE | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[WD = 1 \cdot 553760 \text{ (J)} \quad (= 1553 \cdot 760 \text{ kJ})\] | A1 | cao |
| (b) Using N2L up plane with \(a = 0\): | M1 | All forces, dim. correct, Correct equation |
| \[F - 16 - 80g\left(\frac{2}{7}\right) = 0 \quad F - 240 = 0\] | A1 | \(\frac{250}{v} - 16 - 80g\left(\frac{2}{7}\right) = 0\) |
| \[v = \frac{25}{24} = 1 \cdot 04(166\ldots) \text{ (ms}^{-1}\text{)} \quad \text{(max. speed)}\] | A1 | cao |
| \(F = \frac{250}{v}\) (maximum force) | B1 | si |
(a)(i) Work-energy principle: $16 \times d = 1440(000)$ | M1 | Used, $F \times d = E$
$d = 90(000)$ (km (m)) | A1 | cao, oe
(ii) Using expression for PE or KE: | M1 |
At end, $KE = \frac{1}{2}(80)(10)^2 = 4000$ | A1 | $= 4000$ J $= 4$ kJ
$$PE = 80gh = \begin{cases}
11200g = 109760 & \text{for } h = 140 \\
12000g = 117600 & \text{for } h = 150 \\
800g = 7840 & \text{for } h = 10
\end{cases}$$ | A1 |
Work-energy principle (See notes): All terms included, oe FT their KE and PE | M1 |
$$WD = 1440000 + 4000 + 109760$$
$$(WD = 1440000 + 4000 + 11200g)$$
$$WD = 1 \cdot 553760 \text{ (J)} \quad (= 1553 \cdot 760 \text{ kJ})$$ | A1 | cao
(b) Using N2L up plane with $a = 0$: | M1 | All forces, dim. correct, Correct equation
$$F - 16 - 80g\left(\frac{2}{7}\right) = 0 \quad F - 240 = 0$$ | A1 | $\frac{250}{v} - 16 - 80g\left(\frac{2}{7}\right) = 0$
$$v = \frac{25}{24} = 1 \cdot 04(166\ldots) \text{ (ms}^{-1}\text{)} \quad \text{(max. speed)}$$ | A1 | cao
$F = \frac{250}{v}$ (maximum force) | B1 | si
---Geraint is a cyclist competing in a race along the Taff Trail. The Taff Trail is a track that runs from Cardiff Bay to Brecon. The chart below shows the altitude (height above sea level) along the route.
\includegraphics{figure_4}
Geraint starts from rest at Cardiff Bay and has a speed of $10$ ms$^{-1}$ when he crosses the finish line in Brecon. Geraint and his bike have a total mass of $80$ kg. The resistance to motion may be modelled by a constant force of magnitude $16$ N.
\begin{enumerate}[label=(\alph*)]
\item Given that $1440$ kJ of energy is used in overcoming resistances during the race,
\begin{enumerate}[label=(\roman*)]
\item find the length of the track,
\item calculate the work done by Geraint. [8]
\end{enumerate}
\item The steepest section of the track may be modelled as a slope inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac{2}{7}$.
\includegraphics{figure_4b}
Geraint is capable of producing a maximum power of $250$ W. Find the maximum speed that Geraint can attain whilst travelling on this section of the track. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2023 Q4 [13]}}