| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 9 |
| Topic | Advanced work-energy problems |
| Type | Work done by vector force displacement |
| Difficulty | Standard +0.8 This Further Mechanics question requires differentiation of vector velocity to find force, computing power via dot product, and applying work-energy theorem with kinetic energy calculation. While the calculus is straightforward, it combines multiple concepts (Newton's second law in vector form, power, work-energy) and requires careful algebraic manipulation across three connected parts, placing it moderately above average difficulty. |
| Spec | 3.03d Newton's second law: 2D vectors6.02d Mechanical energy: KE and PE concepts6.02l Power and velocity: P = Fv |
1 A body, $P$, of mass 2 kg moves under the action of a single force $\mathbf { F } \mathrm { N }$. At time $t \mathrm {~s}$, the velocity of the body is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, where
$$\mathbf { v } = \left( t ^ { 2 } - 3 \right) \mathbf { i } + \frac { 5 } { 2 t + 1 } \mathbf { j } \text { for } t \geq 2 .$$
(i) Obtain $\mathbf { F }$ in terms of $t$.\\
(ii) Calculate the rate at which the force $\mathbf { F }$ is working at $t = 4$.\\
(iii) By considering the change in kinetic energy of $P$, calculate the work done by the force $\mathbf { F }$ during the time interval $2 \leq t \leq 4$.
\hfill \mbox{\textit{OCR Further Mechanics 2017 Q1 [9]}}