Edexcel M2 — Question 2 6 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks6
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Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeRough inclined plane work-energy
DifficultyStandard +0.3 This is a straightforward work-energy problem requiring application of the work-energy principle on an inclined plane. Students must calculate change in KE and change in PE, then sum them - a standard M2 exercise with clear given values and direct application of formulas, slightly above average due to the multi-component calculation.
Spec6.02b Calculate work: constant force, resolved component
2. Charlotte, whose mass is 55 kg , is running up a straight hill inclined at \(6 ^ { \circ }\) to the horizontal. She passes two points \(P\) and \(Q , 80\) metres apart, with speeds \(2 \cdot 5 \mathrm {~ms} ^ { - 1 }\) and \(1 \cdot 5 \mathrm {~ms} ^ { - 1 }\) respectively.
Calculate, in J to the nearest whole number, the total work done by Charlotte as she runs from \(P\) to \(Q\).
AnswerMarks Guidance
Gain in K.E. \(= \frac{1}{2} \times 55 \times (2.5^2 - 1.5^2) = 110 \text{ J}\)M1 A1
Gain in P.E. \(= 55g(80 \sin 6°) = 4507 \text{ J}\)M1 A1
Work done \(=\) total energy gain \(= 4617 \text{ J}\)M1 A1 6 marks 
Gain in K.E. $= \frac{1}{2} \times 55 \times (2.5^2 - 1.5^2) = 110 \text{ J}$ | M1 A1 |
Gain in P.E. $= 55g(80 \sin 6°) = 4507 \text{ J}$ | M1 A1 |
Work done $=$ total energy gain $= 4617 \text{ J}$ | M1 A1 | 6 marks
2. Charlotte, whose mass is 55 kg , is running up a straight hill inclined at $6 ^ { \circ }$ to the horizontal. She passes two points $P$ and $Q , 80$ metres apart, with speeds $2 \cdot 5 \mathrm {~ms} ^ { - 1 }$ and $1 \cdot 5 \mathrm {~ms} ^ { - 1 }$ respectively.\\
Calculate, in J to the nearest whole number, the total work done by Charlotte as she runs from $P$ to $Q$.\\

\hfill \mbox{\textit{Edexcel M2  Q2 [6]}}
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