| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 4 |
| Topic | Work done and energy |
| Type | Lifting objects vertically |
| Difficulty | Moderate -0.8 This is a straightforward application of basic work-energy formulas (W = mgh, P = W/t) with no problem-solving required beyond direct substitution. While it's Further Mechanics, the question involves only routine calculations with given values, making it easier than a typical A-level question that would require multi-step reasoning or conceptual understanding. |
| Spec | 6.02a Work done: concept and definition6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| \(65 \times 9.8 \times 78\) awrt 49 700 (J) | M1, A1 [2] | Allow M mark with g omitted. Or 49.7 kJ but not just 49.7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{49700}{90 \times 60}\) 9.20 (W) | M1, A1 [2] | Dividing their value from (i) by a time |
## (i)
$65 \times 9.8 \times 78$ awrt 49 700 (J) | M1, A1 [2] | Allow M mark with g omitted. Or 49.7 kJ but not just 49.7
## (ii)
$\frac{49700}{90 \times 60}$ 9.20 (W) | M1, A1 [2] | Dividing their value from (i) by a time
---1 A climber of mass 65 kg climbs from the bottom to the top of a vertical cliff which is 78 m in height. The climb takes 90 minutes so the velocity of the climber can be neglected.\\
(i) Calculate the work done by the climber in climbing the cliff.\\
(ii) Calculate the average power generated by the climber in climbing the cliff.
\hfill \mbox{\textit{OCR FM1 AS 2017 Q1 [4]}}