| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Lifting objects vertically |
| Difficulty | Easy -1.2 This is a straightforward two-part question requiring only direct application of basic formulas: T = mg for constant speed, then P = W/t where W = mgh. No problem-solving insight needed, just routine substitution into standard mechanics formulas with clearly given values. |
| Spec | 3.03f Weight: W=mg6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Tension is 8000 N or 800\(g\) | B1 | Accept 7840 N (from 9.8) or 7850 (from 9.81) |
| Answer | Marks | Guidance |
|---|---|---|
| For using \(P = \frac{\Delta W}{\Delta t}\) or \(P = Tv\) | M1 | |
| \(\Delta W = 8000 \times 20\) or \(v = \frac{20}{50}\) | A1ft | |
| Power applied is 3200 W | A1 | Accept 3140 W (from 9.8 or 9.81) |
# Question 1:
## Part (i)
| Tension is 8000 N or 800$g$ | B1 | Accept 7840 N (from 9.8) or 7850 (from 9.81) |
## Part (ii)
| For using $P = \frac{\Delta W}{\Delta t}$ or $P = Tv$ | M1 | |
| $\Delta W = 8000 \times 20$ or $v = \frac{20}{50}$ | A1ft | |
| Power applied is 3200 W | A1 | Accept 3140 W (from 9.8 or 9.81) |
**SR** (for candidates who omit $g$) (Max 2 out of 3):
$P = 800 \times 20 \div 50$ B1, Power applied is 320 W B1
---1 A crate of mass 800 kg is lifted vertically, at constant speed, by the cable of a crane. Find\\
(i) the tension in the cable,\\
(ii) the power applied to the crate in increasing the height by 20 m in 50 s .
\hfill \mbox{\textit{CAIE M1 2003 Q1 [4]}}