CAIE M1 2012 November — Question 1 3 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2012
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeWork done by constant force - find work done
DifficultyEasy -1.3 This is a straightforward application of the work formula W = Fs cos θ with all values given directly. It requires only substitution into a single formula (45 × 25 × cos 14°) with no problem-solving, making it significantly easier than average and purely procedural.
Spec6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component
1 \includegraphics[max width=\textwidth, alt={}, center]{9fbb63e3-4017-461e-9110-500be2c20778-2_122_803_255_671} A block is pushed along a horizontal floor by a force of magnitude 45 N acting at an angle of \(14 ^ { \circ }\) to the horizontal (see diagram). Find the work done by the force in moving the block a distance of 25 m .
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(WD = 45 \times 25\cos 14°\)M1 For using \(WD = Fd\cos\alpha\)
\(WD = 45 \times 25\cos 14°\)A1
Work done is 1090 J (1.09 kJ)A1 [3] 
# Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $WD = 45 \times 25\cos 14°$ | M1 | For using $WD = Fd\cos\alpha$ |
| $WD = 45 \times 25\cos 14°$ | A1 | |
| Work done is 1090 J (1.09 kJ) | A1 | **[3]** |

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1\\
\includegraphics[max width=\textwidth, alt={}, center]{9fbb63e3-4017-461e-9110-500be2c20778-2_122_803_255_671}

A block is pushed along a horizontal floor by a force of magnitude 45 N acting at an angle of $14 ^ { \circ }$ to the horizontal (see diagram). Find the work done by the force in moving the block a distance of 25 m .

\hfill \mbox{\textit{CAIE M1 2012 Q1 [3]}}
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