| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Cyclist or runner: find resistance or speed |
| Difficulty | Standard +0.3 This is a standard M2 power and work-energy question requiring straightforward application of F=ma to find driving force, then P=Fv, followed by routine use of work-energy principle. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks |
|---|---|
| Driving force = \(F\) Resolving parallel to the plane: \(F - 20 - 75g\sin 5 = -75 \times 0.2 = -15\) | M1 A2 - lee |
| \(F = 5 + 75g\sin 5°\) | |
| \(P = Fv\) \(\therefore\) working at \(12 \times (5 + 75g\sin 5°) = 828.7...\approx 830\) W | DM1 A1 |
| Answer | Marks |
|---|---|
| Loss in KE = gain in GPE + work done against resistance | M1 |
| \(\frac{1}{2} \times 75 \times 64 = 75 \times 9.8 \times \sin 5° d + 20d \times 84.059....\) | A2 - lee |
| \(d = 28.6\) m | DM1 A1 |
## (a)
| Driving force = $F$ Resolving parallel to the plane: $F - 20 - 75g\sin 5 = -75 \times 0.2 = -15$ | M1 A2 - lee |
| $F = 5 + 75g\sin 5°$ | |
| $P = Fv$ $\therefore$ working at $12 \times (5 + 75g\sin 5°) = 828.7...\approx 830$ W | DM1 A1 |
**Total: 5 marks**
## (b)
| Loss in KE = gain in GPE + work done against resistance | M1 |
| $\frac{1}{2} \times 75 \times 64 = 75 \times 9.8 \times \sin 5° d + 20d \times 84.059....$ | A2 - lee |
| $d = 28.6$ m | DM1 A1 |
**Total: 5 marks**
**Question 3 Total: 10 marks**
---A cyclist and her cycle have a combined mass of $75$ kg. The cyclist is cycling up a straight road inclined at $5°$ to the horizontal. The resistance to the motion of the cyclist from non-gravitational forces is modelled as a constant force of magnitude $20$ N. At the instant when the cyclist has a speed of $12$ m s$^{-1}$, she is decelerating at $0.2$ m s$^{-2}$.
\begin{enumerate}[label=(\alph*)]
\item Find the rate at which the cyclist is working at this instant.
[5]
\end{enumerate}
When the cyclist passes the point $A$ her speed is $8$ m s$^{-1}$. At $A$ she stops working but does not apply the brakes. She comes to rest at the point $B$.
The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude $20$ N.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use the work-energy principle to find the distance $AB$.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q3 [10]}}