CAIE M1 2008 November — Question 4 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2008
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeLifting objects vertically
DifficultyModerate -0.8 This is a straightforward application of the work-energy theorem requiring only basic kinematics (finding displacement from constant acceleration) and the work-energy principle. The question guides students by explicitly stating 'by considering energy,' making it easier than average with no conceptual challenges beyond standard M1 content.
Spec6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae

4 A load of mass 160 kg is lifted vertically by a crane, with constant acceleration. The load starts from rest at the point \(O\). After 7 s , it passes through the point \(A\) with speed \(0.5 \mathrm {~ms} ^ { - 1 }\). By considering energy, find the work done by the crane in moving the load from \(O\) to \(A\).

Question 4:
AnswerMarks Guidance
\([s = (0 + 0.5)/2 \times 7]\)M1 For using \((u+v)/2 = s/t\)
\(s = 1.75\) mA1 May be implied
PE gain \(= 160g \times 1.75\)B1ft
KE gain \(= \frac{1}{2} \times 160 \times 0.5^2\)B1
\([WD = 2800 + 20]\)M1 For using \(WD =\) PE gain \(+\) KE gain
Work done is \(2820\) JA1 [6]
SR (max 4/6) for candidates who use a non-energy method:
AnswerMarks Guidance
\([s = (0 + 0.5)/2 \times 7]\)M1 For using \((u+v)/2 = s/t\)
\(s = 1.75\) mA1
\([a = 1/14,\ T = 160g + 160/14,\ WD = 1611.4... \times 1.75]\)M1 For finding the acceleration and using Newton's second law (3 terms) to find the tension in the rope, then multiplying by the distance
Work done is \(2820\) JA1
# Question 4:

$[s = (0 + 0.5)/2 \times 7]$ | M1 | For using $(u+v)/2 = s/t$
$s = 1.75$ m | A1 | May be implied
PE gain $= 160g \times 1.75$ | B1ft |
KE gain $= \frac{1}{2} \times 160 \times 0.5^2$ | B1 |
$[WD = 2800 + 20]$ | M1 | For using $WD =$ PE gain $+$ KE gain
Work done is $2820$ J | A1 | [6]

**SR (max 4/6) for candidates who use a non-energy method:**
$[s = (0 + 0.5)/2 \times 7]$ | M1 | For using $(u+v)/2 = s/t$
$s = 1.75$ m | A1 |
$[a = 1/14,\ T = 160g + 160/14,\ WD = 1611.4... \times 1.75]$ | M1 | For finding the acceleration and using Newton's second law (3 terms) to find the tension in the rope, then multiplying by the distance
Work done is $2820$ J | A1 |

---
4 A load of mass 160 kg is lifted vertically by a crane, with constant acceleration. The load starts from rest at the point $O$. After 7 s , it passes through the point $A$ with speed $0.5 \mathrm {~ms} ^ { - 1 }$. By considering energy, find the work done by the crane in moving the load from $O$ to $A$.

\hfill \mbox{\textit{CAIE M1 2008 Q4 [6]}}