Edexcel M2 2011 June — Question 1 5 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2011
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicWork done and energy
TypeConstant speed up/down incline
DifficultyModerate -0.3 This is a straightforward application of the power equation P = Fv at constant speed, where driving force equals resistance plus component of weight down the slope. The calculation involves standard mechanics formulas with no conceptual difficulty—slightly easier than average due to being a single-concept, direct application problem with given numerical values.
Spec6.02k Power: rate of doing work6.02l Power and velocity: P = Fv
A car of mass 1000 kg moves with constant speed \(V\) m s\(^{-1}\) up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac{1}{30}\). The engine of the car is working at a rate of 12 kW. The resistance to motion from non-gravitational forces has magnitude 500 N. Find the value of \(V\). [5]
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(12000 = TV\)M1
\(T - 500 - 1000 g \sin \theta = 0\)M1 A1
\(V = \frac{12000}{500 + 1000 \times 9.8 \times \frac{s}{d}}\)
\(V = 15\) (accept 14.5)DM1 A1 
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $12000 = TV$ | M1 | |
| $T - 500 - 1000 g \sin \theta = 0$ | M1 A1 | |
| $V = \frac{12000}{500 + 1000 \times 9.8 \times \frac{s}{d}}$ | | |
| $V = 15$ (accept 14.5) | DM1 A1 | |
A car of mass 1000 kg moves with constant speed $V$ m s$^{-1}$ up a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{1}{30}$. The engine of the car is working at a rate of 12 kW. The resistance to motion from non-gravitational forces has magnitude 500 N.

Find the value of $V$. [5]

\hfill \mbox{\textit{Edexcel M2 2011 Q1 [5]}}
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Q1 5 Q2 8 Q3 8 Q4 7 Q5 10 Q6 11 Q7 13 Q8 13