| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on horizontal road |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of friction force (F = μR), work-energy principle, and kinematics. Part (a) involves simple calculation of friction work (μmg × distance), and part (b) applies work-energy theorem with given initial speed. All steps are routine textbook procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02c Work by variable force: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R(\uparrow): R = 10g\) | B1 | |
| \(F = \mu R \Rightarrow F = \dfrac{4}{7}(10g) = 56\) | B1 | |
| \(\therefore\) WD against friction \(= \dfrac{4}{7}(10g)(50)\) | M1 | |
| \(2800\) (J) | A1 | |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(70(50) - \text{"2800"} = \frac{1}{2}(10)v^2 - \frac{1}{2}(10)(2)^2\) | M1*, A1ft | |
| \(700 = 5v^2 - 20,\ 5v^2 = 720 \Rightarrow v^2 = 144\) | d*M1 | |
| Hence \(v = \underline{12}\) (m s\(^{-1}\)) | A1 cao | |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| N2L\((\rightarrow)\): \(70 - \dfrac{4}{7}R = 10a\) | M1* | |
| \(70 - \dfrac{4}{7} \times 10g = 10a,\ (a = 1.4)\) | A1ft | |
| \(AB(\rightarrow): v^2 = (2)^2 + 2(1.4)(50)\) | d*M1 | |
| Hence \(v = \underline{12}\) (m s\(^{-1}\)) | A1 cao | |
| Total: [8] |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R(\uparrow): R = 10g$ | B1 | |
| $F = \mu R \Rightarrow F = \dfrac{4}{7}(10g) = 56$ | B1 | |
| $\therefore$ WD against friction $= \dfrac{4}{7}(10g)(50)$ | M1 | |
| $2800$ (J) | A1 | |
| **Total: (4)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $70(50) - \text{"2800"} = \frac{1}{2}(10)v^2 - \frac{1}{2}(10)(2)^2$ | M1*, A1ft | |
| $700 = 5v^2 - 20,\ 5v^2 = 720 \Rightarrow v^2 = 144$ | d*M1 | |
| Hence $v = \underline{12}$ (m s$^{-1}$) | A1 cao | |
| **Total: (4)** | | |
### Part (b) – Alternative Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| N2L$(\rightarrow)$: $70 - \dfrac{4}{7}R = 10a$ | M1* | |
| $70 - \dfrac{4}{7} \times 10g = 10a,\ (a = 1.4)$ | A1ft | |
| $AB(\rightarrow): v^2 = (2)^2 + 2(1.4)(50)$ | d*M1 | |
| Hence $v = \underline{12}$ (m s$^{-1}$) | A1 cao | |
| **Total: [8]** | | |
---\begin{enumerate}
\item A block of mass 10 kg is pulled along a straight horizontal road by a constant horizontal force of magnitude 70 N in the direction of the road. The block moves in a straight line passing through two points $A$ and $B$ on the road, where $A B = 50 \mathrm {~m}$. The block is modelled as a particle and the road is modelled as a rough plane. The coefficient of friction between the block and the road is $\frac { 4 } { 7 }$.\\
(a) Calculate the work done against friction in moving the block from $A$ to $B$.
\end{enumerate}
The block passes through $A$ with a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(b) Find the speed of the block at $B$.\\
\hfill \mbox{\textit{Edexcel M2 2009 Q3 [8]}}