| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Easy -1.3 This is a straightforward application of basic work-energy formulas: W = Fs cos θ for part (i) and P = Fv cos θ for part (ii). Both parts require only direct substitution into standard formulas with no problem-solving or conceptual challenges, making it easier than average. |
| Spec | 6.02a Work done: concept and definition6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(WD = 65 \times 76\cos 5°\) | M1 | For using \(WD = T d\cos \alpha\) |
| Work done is 4920 J | A1 | [2] |
| (ii) \(P = 65 \cos 5° \times 1.5\) | M1 | For using \(P = Tv\cos \alpha\) |
| Rate of working is 97.1 W | A1, B1 | [2] |
**(i)** $WD = 65 \times 76\cos 5°$ | M1 | For using $WD = T d\cos \alpha$
Work done is 4920 J | A1 | [2]
**(ii)** $P = 65 \cos 5° \times 1.5$ | M1 | For using $P = Tv\cos \alpha$
Rate of working is 97.1 W | A1, B1 | [2] | $ft$ for the value of $\text{ans(i)} \times 1.5 = 76$ SR for candidates who assume without justification that the speed is constant (max 1/2) $t = 76 \div 1.5 = 50.6...s$ rate $= WD/t = 4960 \div 50.6... = 97.1W$1 A load is pulled along horizontal ground for a distance of 76 m , using a rope. The rope is inclined at $5 ^ { \circ }$ above the horizontal and the tension in the rope is 65 N .\\
(i) Find the work done by the tension.
At an instant during the motion the velocity of the load is $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the rate of working of the tension at this instant.
\hfill \mbox{\textit{CAIE M1 2011 Q1 [4]}}