| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Lifting objects vertically |
| Difficulty | Moderate -0.3 This is a straightforward multi-part work-energy question requiring standard applications of formulas (W = mgh + ½mv², P = Fv, work-energy principle). All parts involve direct substitution into well-known equations with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Work-energy theorem: \(W - mgh = \frac{1}{2}mv^2 - 0\) | M1 | WE equation with all terms |
| \(W = \frac{1}{2}(20)(0.5)^2 + 20 \times 10 \times 4\) | A1 | |
| \(W = 2.5 + 800 = 802.5\) J | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tension \(T = mg = 20 \times 10 = 200\) N (steady speed) | B1 | |
| Power \(= Tv = 200 \times 0.5 = 100\) W | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| WE: \(mgh - W_f = \frac{1}{2}m(v_2^2 - v_1^2)\) | M1 | |
| \(35 \times 10 \times 3 - W_f = \frac{1}{2}(35)(9-1)\) | A1 | |
| \(1050 - W_f = 140\) | A1 | |
| \(W_f = 910\) J | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Net work done \(=\) change in KE | M1 | |
| \((35 \times 10 \times 0.1 - 150)d = 0 - \frac{1}{2}(35)(9)\) | M1 A1 | Using \(\sin\alpha = 0.1\) |
| \((35 - 150)d = -157.5\) | A1 | |
| \(-115d = -157.5\) | ||
| \(d = \frac{157.5}{115} \approx 1.37\) m | A1 |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Work-energy theorem: $W - mgh = \frac{1}{2}mv^2 - 0$ | M1 | WE equation with all terms |
| $W = \frac{1}{2}(20)(0.5)^2 + 20 \times 10 \times 4$ | A1 | |
| $W = 2.5 + 800 = 802.5$ J | A1 A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tension $T = mg = 20 \times 10 = 200$ N (steady speed) | B1 | |
| Power $= Tv = 200 \times 0.5 = 100$ W | M1 A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| WE: $mgh - W_f = \frac{1}{2}m(v_2^2 - v_1^2)$ | M1 | |
| $35 \times 10 \times 3 - W_f = \frac{1}{2}(35)(9-1)$ | A1 | |
| $1050 - W_f = 140$ | A1 | |
| $W_f = 910$ J | A1 A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Net work done $=$ change in KE | M1 | |
| $(35 \times 10 \times 0.1 - 150)d = 0 - \frac{1}{2}(35)(9)$ | M1 A1 | Using $\sin\alpha = 0.1$ |
| $(35 - 150)d = -157.5$ | A1 | |
| $-115d = -157.5$ | | |
| $d = \frac{157.5}{115} \approx 1.37$ m | A1 | |4 Jack and Jill are raising a pail of water vertically using a light inextensible rope. The pail and water have total mass 20 kg .
In parts (i) and (ii), all non-gravitational resistances to motion may be neglected.\\
(i) How much work is done to raise the pail from rest so that it is travelling upwards at $0.5 \mathrm {~ms} ^ { - 1 }$ when at a distance of 4 m above its starting position?\\
(ii) What power is required to raise the pail at a steady speed of $0.5 \mathrm {~ms} ^ { - 1 }$ ?
Jack falls over and hurts himself. He then slides down a hill.\\
His mass is 35 kg and his speed increases from $1 \mathrm {~ms} ^ { - 1 }$ to $3 \mathrm {~ms} ^ { - 1 }$ while descending through a vertical height of 3 m .\\
(iii) How much work is done against friction?
In Jack's further motion, he slides down a slope at an angle $\alpha$ to the horizontal where $\sin \alpha = 0.1$. The frictional force on him is now constant at 150 N . For this part of the motion, Jack's initial speed is $3 \mathrm {~ms} ^ { - 1 }$.\\
(iv) How much further does he slide before coming to rest?
\hfill \mbox{\textit{OCR MEI M2 2007 Q4 [17]}}