| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Work done by vector force displacement |
| Difficulty | Moderate -0.3 This is a straightforward M2 work-energy question requiring standard application of formulas: work = force × distance × cos(angle), PE = mgh, and energy conservation. All values are given directly, requiring only careful calculation with no problem-solving insight or novel approach needed. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(WD = 100\cos20° \times 30\) | M1 | Product of 3 relevant elements. Angle could be 5, 25 or complements. \(2819.1\)... |
| \(WD = 2820\) J | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(PE = 25g \times 30\sin5\) | M1 | Product of weight and vertical height. Allow without \(g\). \(640.6\) |
| \(PE = 641\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2819.1 = 640.6 + 30x70 + 25v^2/2\) | M1 | 4 term energy equation. \(ft(cv\) 2820 and \(cv\) 641\()\) |
| \(v = 2.51\) m s\(^{-1}\) | A1ft | |
| A1 | cao | |
| [4] | ||
| OR \(25a = 100\cos20 -70 -25g\sin5\) | *M1 | 4 term equation. Allow 0.1 here. Or equivalent complete method. |
| \(a = 0.105\) | A1 | cao |
| \(v^2 = 2 \times 30 \times 'a'\) | dep*M1 | |
| \(v = 2.51\) | A1 | cao |
| [4] |
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $WD = 100\cos20° \times 30$ | M1 | Product of 3 relevant elements. Angle could be 5, 25 or complements. $2819.1$... |
| $WD = 2820$ J | A1 | |
| | **[2]** | |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $PE = 25g \times 30\sin5$ | M1 | Product of weight and vertical height. Allow without $g$. $640.6$ |
| $PE = 641$ | A1 | |
| | **[2]** | |
**Part (iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2819.1 = 640.6 + 30x70 + 25v^2/2$ | M1 | 4 term energy equation. $ft(cv$ 2820 and $cv$ 641$)$ |
| $v = 2.51$ m s$^{-1}$ | A1ft | |
| | A1 | cao |
| | **[4]** | |
| OR $25a = 100\cos20 -70 -25g\sin5$ | *M1 | 4 term equation. Allow 0.1 here. Or equivalent complete method. |
| $a = 0.105$ | A1 | cao |
| $v^2 = 2 \times 30 \times 'a'$ | dep*M1 | |
| $v = 2.51$ | A1 | cao |
| | **[4]** | |4 A block of mass 25 kg is dragged 30 m up a slope inclined at $5 ^ { \circ }$ to the horizontal by a rope inclined at $20 ^ { \circ }$ to the slope. The tension in the rope is 100 N and the resistance to the motion of the block is 70 N . The block is initially at rest. Calculate\\
(i) the work done by the tension in the rope,\\
(ii) the change in the potential energy of the block,\\
(iii) the speed of the block after it has moved 30 m up the slope.
\hfill \mbox{\textit{OCR M2 2011 Q4 [8]}}