| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find steady/maximum speed given power |
| Difficulty | Moderate -0.3 This is a straightforward application of the power formula P=Fv with constant speed (so net force = 0). Part (a) requires simple multiplication, part (b) requires resolving forces on an incline and solving a linear equation. Standard M2 bookwork with no problem-solving insight needed, slightly easier than average. |
| Spec | 3.03f Weight: W=mg3.03g Gravitational acceleration6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Constant \(v \Rightarrow\) driving force \(=\) resistance \(\Rightarrow F = 120\) (N) | M1 | |
| \(\Rightarrow P = 120 \times 10 = 1200\text{W}\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving parallel to slope, zero acceleration: \(\frac{P}{v} = 120 + 300g\sin\theta\ (= 330)\) | M1A1A1 | |
| \(\Rightarrow v = \frac{1200}{330} = 3.6 \text{ (ms}^{-1})\) | A1 |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Constant $v \Rightarrow$ driving force $=$ resistance $\Rightarrow F = 120$ (N) | M1 | |
| $\Rightarrow P = 120 \times 10 = 1200\text{W}$ | M1 | |
**Subtotal: (2)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving parallel to slope, zero acceleration: $\frac{P}{v} = 120 + 300g\sin\theta\ (= 330)$ | M1A1A1 | |
| $\Rightarrow v = \frac{1200}{330} = 3.6 \text{ (ms}^{-1})$ | A1 | |
**Subtotal: (4) Total: [6]**
---3. A truck of mass of 300 kg moves along a straight horizontal road with a constant speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The resistance to motion of the truck has magnitude 120 N .
\begin{enumerate}[label=(\alph*)]
\item Find the rate at which the engine of the truck is working.
On another occasion the truck moves at a constant speed up a hill inclined at $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 14 }$. The resistance to motion of the truck from non-gravitational forces remains of magnitude 120 N . The rate at which the engine works is the same as in part (a).
\item Find the speed of the truck.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q3 [6]}}