| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2004 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Work done by vector force displacement |
| Difficulty | Standard +0.3 This is a straightforward M5 mechanics question requiring standard application of Newton's second law in 3D and work-energy calculations. Students must find acceleration from kinematics (using s=ut+½at²), apply F=ma to find F₂, then calculate work using W=F·s. All steps are routine for Further Maths students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication3.02d Constant acceleration: SUVAT formulae6.02b Calculate work: constant force, resolved component |
2. Two constant forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ are the only forces acting on a particle $P$ of mass 2 kg . The particle is initially at rest at the point $A$ with position vector $( - 2 \mathbf { i } - \mathbf { j } - 4 \mathbf { k } ) \mathrm { m }$. Four seconds later, $P$ is at the point $B$ with position vector $( 6 \mathbf { i } - 5 \mathbf { j } + 8 \mathbf { k } ) \mathrm { m }$.
Given that $\mathbf { F } _ { 1 } = ( 12 \mathbf { i } - 4 \mathbf { j } + 6 \mathbf { k } ) \mathrm { N }$, find
\begin{enumerate}[label=(\alph*)]
\item $\mathbf { F } _ { 2 }$,
\item the work done on $P$ as it moves from $A$ to $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2004 Q2 [8]}}