| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Cyclist or runner: find resistance or speed |
| Difficulty | Moderate -0.3 This is a straightforward power calculation question requiring the formula P = Fv. Part (a) is direct substitution, while part (b) requires resolving forces on an incline to find the driving force, then applying the same formula. The incline component is standard FM1 material with a convenient sin value given, making calculations simple. Slightly below average difficulty as it's a routine application of well-practiced techniques with no problem-solving insight required. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Force = Resistance (since no acceleration) \(= 30\) | B1 | For force \(= 30\) seen |
| Power \(= \text{Force} \times \text{Speed} = 30 \times 4\) | M1 | For use of \(P = Fv\) |
| \(= 120 \text{ W}\) | A1ft | For 120 (W), follow through on their '30' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolving parallel to the slope | M1 | For resolving parallel to slope with correct no. of terms and \(60g\) resolved |
| \(F - 60g\sin\alpha - 30 = 0\) | A1 | For a correct equation |
| \(F = 70\) | A1 | For \(F = 70\) |
| Power \(= \text{Force} \times \text{Speed} = 70 \times 3\) | M1 | For use of \(P = Fv\) |
| \(= 210 \text{ W}\) | A1ft | For 210 (W), follow through on their '70' |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Force = Resistance (since no acceleration) $= 30$ | B1 | For force $= 30$ seen |
| Power $= \text{Force} \times \text{Speed} = 30 \times 4$ | M1 | For use of $P = Fv$ |
| $= 120 \text{ W}$ | A1ft | For 120 (W), follow through on their '30' |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving parallel to the slope | M1 | For resolving parallel to slope with correct no. of terms and $60g$ resolved |
| $F - 60g\sin\alpha - 30 = 0$ | A1 | For a correct equation |
| $F = 70$ | A1 | For $F = 70$ |
| Power $= \text{Force} \times \text{Speed} = 70 \times 3$ | M1 | For use of $P = Fv$ |
| $= 210 \text{ W}$ | A1ft | For 210 (W), follow through on their '70' |
---\begin{enumerate}
\item \hspace{0pt} [In this question use $\mathrm { g } = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ ]
\end{enumerate}
A jogger of mass 60 kg runs along a straight horizontal road at a constant speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The total resistance to the motion of the jogger is modelled as a constant force of magnitude 30 N .\\
(a) Find the rate at which the jogger is working.
The jogger now comes to a hill which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 1 } { 15 }$. Because of the hill, the jogger reduces her speed to $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and maintains this constant speed as she runs up the hill. The total resistance to the motion of the jogger from non-gravitational forces continues to be modelled as a constant force of magnitude 30 N .\\
(b) Find the rate at which she has to work in order to run up the hill at $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{Edexcel FM1 AS Q3 [8]}}