| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 work-energy question with friction on an inclined plane. Part (a) is straightforward PE calculation, (b) applies work-energy principle directly, (c) uses standard friction formula, and (d) repeats the method with different initial conditions. All steps follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| PE lost \(= 3 \times g \times 8\sin 30 = 3 \times g \times 8 \times 0.5 = 117.6 \text{ J} \approx 118 \text{ J or } 120 \text{ J}\) | M1 A1 | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| KE gained \(= \frac{1}{2} \times 3 \times 5^2 = 37.5 \text{ J}\) | M1 A1 | |
| Work-energy: \(F \times 8 = 117.6 - 37.5 = 80.1\) | M1 A1 \(\checkmark\) | |
| \(\Rightarrow F = 10.0125 \approx 10 \text{ N}\) | A1 | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(R = 3g\cos 30\ (= 25.46 \text{ N})\) | B1 | |
| \(F = \mu R \Rightarrow \mu = \frac{10}{25.46} \approx 0.393 \text{ or } 0.39\) | M1 A1 | |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Work done by friction \(= 80.1\) as before | M1 | |
| Work-energy: \(\frac{1}{2} \times 3 \times v^2 = \frac{1}{2} \times 3 \times 2^2 + 117.6 - 80.1\) | M1 A2,1,0 \(\checkmark\) | |
| \(\Rightarrow v \approx 5.39 \text{ or } 5.4 \text{ ms}^{-1}\) | A1 | |
| (5) |
## Question 7:
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| PE lost $= 3 \times g \times 8\sin 30 = 3 \times g \times 8 \times 0.5 = 117.6 \text{ J} \approx 118 \text{ J or } 120 \text{ J}$ | M1 A1 | |
| | **(2)** | |
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| KE gained $= \frac{1}{2} \times 3 \times 5^2 = 37.5 \text{ J}$ | M1 A1 | |
| Work-energy: $F \times 8 = 117.6 - 37.5 = 80.1$ | M1 A1 $\checkmark$ | |
| $\Rightarrow F = 10.0125 \approx 10 \text{ N}$ | A1 | |
| | **(5)** | |
### Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $R = 3g\cos 30\ (= 25.46 \text{ N})$ | B1 | |
| $F = \mu R \Rightarrow \mu = \frac{10}{25.46} \approx 0.393 \text{ or } 0.39$ | M1 A1 | |
| | **(3)** | |
### Part (d):
| Working | Marks | Notes |
|---------|-------|-------|
| Work done by friction $= 80.1$ as before | M1 | |
| Work-energy: $\frac{1}{2} \times 3 \times v^2 = \frac{1}{2} \times 3 \times 2^2 + 117.6 - 80.1$ | M1 A2,1,0 $\checkmark$ | |
| $\Rightarrow v \approx 5.39 \text{ or } 5.4 \text{ ms}^{-1}$ | A1 | |
| | **(5)** | |7. At a demolition site, bricks slide down a straight chute into a container. The chute is rough and is inclined at an angle of $30 ^ { \circ }$ to the horizontal. The distance travelled down the chute by each brick is 8 m . A brick of mass 3 kg is released from rest at the top of the chute. When it reaches the bottom of the chute, its speed is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the potential energy lost by the brick in moving down the chute.
\item By using the work-energy principle, or otherwise, find the constant frictional force acting on the brick as it moves down the chute.
\item Hence find the coefficient of friction between the brick and the chute.
Another brick of mass 3 kg slides down the chute. This brick is given an initial speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the top of the chute.
\item Find the speed of this brick when it reaches the bottom of the chute.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2005 Q7 [15]}}