| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Topic | Power and driving force |
| Type | Power from work done over time (P = W/t) |
| Difficulty | Standard +0.3 This is a straightforward power calculation requiring students to find work done lifting water (mgh) and kinetic energy from flow velocity, then divide by time. The steps are standard and mechanical with no conceptual subtleties, making it slightly easier than average despite being from Further Maths. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
Work against gravity: $3 \times 10^3 \times 25 \times 10 = 750\,000$ J **M1A1**
Let speed of delivery be $v$ ms$^{-1}$.
$\pi \times 0.05^2 \times 60v = 3$ **M1A1**
$\Rightarrow v = \frac{3}{\pi \times 0.05^2 \times 60} = 6.366$ or $\frac{20}{\pi}$ **A1**
Kinetic energy $\frac{1}{2} \times 3 \times 10^3 \times 6.366^2 = (60792)$ **B1**
Power $= \frac{750000 + 60792}{60 \times 1000} = 13.5$ kW **M1A1** [8]7 Find the power required to pump $3 \mathrm {~m} ^ { 3 }$ of water per minute from a depth of 25 m and deliver it through a circular pipe of diameter 10 cm . Assume that friction may be neglected and that the density of water is $1000 \mathrm {~kg} \mathrm {~m} ^ { - 3 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q7 [8]}}