| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Lifting objects vertically |
| Difficulty | Moderate -0.3 This is a straightforward M2 work-energy-power question requiring standard formulas (PE = mgh, KE = ½mv², Power = Energy/time) with given values. The main steps are calculating mass flow rate from pipe dimensions, then applying energy formulas—routine mechanics with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Volume per second \(= 12\pi(0.04)^2 = 0.0603 \text{ m}^3\), Mass \(= 60.3 \text{ kg}\) | M1 A1 M1 A1 | |
| P.E. gained per sec. \(= 60.3 \times g \times 25 = 14778 \text{ J}\) | A1 | |
| K.E. gained per sec. \(= \frac{1}{2} \times 60.3 \times 12^2 = 4342 \text{ J}\) | M1 A1 | |
| (b) Power \(= \text{total energy per second} = 19120 \text{ J s}^{-1} = 19.1 \text{ kW}\) | M1 A1 | Total: 7 marks |
(a) Volume per second $= 12\pi(0.04)^2 = 0.0603 \text{ m}^3$, Mass $= 60.3 \text{ kg}$ | M1 A1 M1 A1 |
P.E. gained per sec. $= 60.3 \times g \times 25 = 14778 \text{ J}$ | A1 |
K.E. gained per sec. $= \frac{1}{2} \times 60.3 \times 12^2 = 4342 \text{ J}$ | M1 A1 |
(b) Power $= \text{total energy per second} = 19120 \text{ J s}^{-1} = 19.1 \text{ kW}$ | M1 A1 | Total: 7 marksA pump raises water from a reservoir at a depth of 25 m below ground level. The water is delivered at ground level with speed 12 ms$^{-1}$ through a pipe of radius 4 cm. Find
\begin{enumerate}[label=(\alph*)]
\item the potential and kinetic energy given to the water each second, [5 marks]
\item the rate, in kW, at which the pump is working. [2 marks]
\end{enumerate}
[1 m$^3$ of water has a mass of 1000 kg.]
\hfill \mbox{\textit{Edexcel M2 Q2 [7]}}