Edexcel M2 — Question 2 17 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks17
PaperDownload PDF ↗
TopicWork done and energy
TypeAcceleration from power and speed
DifficultyStandard +0.3 This is a standard M2 question combining power-force-acceleration relationships and work-energy principle. Part (a) uses P=Fv with F=ma+R (4 marks), part (b) applies work-energy directly (3 marks), and part (c) is a simple modeling comment (1 mark). All techniques are routine M2 content with straightforward multi-step calculations requiring no novel insight.
Spec3.03d Newton's second law: 2D vectors3.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.04b Equilibrium: zero resultant moment and force6.02a Work done: concept and definition6.02l Power and velocity: P = Fv

A car of mass 1000 kg is moving along a straight horizontal road with a constant acceleration of \(j\) m s\(^{-2}\). The resistance to motion is modelled as a constant force of magnitude 1200 N. When the car is travelling at 12 m s\(^{-1}\), the power generated by the engine of the car is 24 kW.
  1. Calculate the value of \(j\). [4]
When the car is travelling at 14 m s\(^{-1}\), the engine is switched off and the car comes to rest, without braking, in a distance of \(d\) metres. Assuming the same model for resistance,
  1. use the work-energy principle to calculate the value of \(d\). [3]
  1. Give a reason why the model used for the resistance to motion may not be realistic. [1]
A uniform ladder \(AB\), of mass \(m\) and length \(2a\), has one end \(A\) on rough horizontal ground. The other end \(B\) rests against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall. The ladder makes an angle \(α\) with the horizontal, where \(\tan α = \frac{4}{3}\). A child of mass \(2m\) stands on the ladder at \(C\) where \(AC = \frac{1}{4}a\), as shown in Fig. 1. The ladder and the child are in equilibrium. By modelling the ladder as a rod and the child as a particle, calculate the least possible value of the coefficient of friction between the ladder and the ground. [9]

A car of mass 1000 kg is moving along a straight horizontal road with a constant acceleration of $j$ m s$^{-2}$. The resistance to motion is modelled as a constant force of magnitude 1200 N. When the car is travelling at 12 m s$^{-1}$, the power generated by the engine of the car is 24 kW.

\begin{enumerate}[label=(\alph*)]
\item Calculate the value of $j$.
[4]
\end{enumerate}

When the car is travelling at 14 m s$^{-1}$, the engine is switched off and the car comes to rest, without braking, in a distance of $d$ metres. Assuming the same model for resistance,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item use the work-energy principle to calculate the value of $d$.
[3]
\end{enumerate}

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Give a reason why the model used for the resistance to motion may not be realistic.
[1]
\end{enumerate}

A uniform ladder $AB$, of mass $m$ and length $2a$, has one end $A$ on rough horizontal ground. The other end $B$ rests against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall. The ladder makes an angle $α$ with the horizontal, where $\tan α = \frac{4}{3}$. A child of mass $2m$ stands on the ladder at $C$ where $AC = \frac{1}{4}a$, as shown in Fig. 1. The ladder and the child are in equilibrium.

By modelling the ladder as a rod and the child as a particle, calculate the least possible value of the coefficient of friction between the ladder and the ground.
[9]

\hfill \mbox{\textit{Edexcel M2  Q2 [17]}}