AQA Further Paper 3 Mechanics 2020 June — Question 7 8 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2020
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - driving force on horizontal road
DifficultyStandard +0.3 This is a standard work-energy problem requiring resolution of forces, calculation of friction, and application of the work-energy principle. Part (a) involves routine mechanics techniques (resolving forces, finding normal reaction, calculating work done by tension and friction, applying KE change). Part (b) requires basic modelling awareness. While it's a multi-step problem worth several marks, all techniques are standard for Further Maths mechanics with no novel insight required, making it slightly easier than average.
Spec6.02i Conservation of energy: mechanical energy principle

7 In this question use \(g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) A box, of mass 8 kg , is on a rough horizontal surface.
A string attached to the box is used to pull it along the surface.
The string is inclined at an angle of \(40 ^ { \circ }\) above the horizontal.
The tension in the string is 50 newtons.
As the box moves a distance of \(x\) metres, its speed increases from \(2 \mathrm {~ms} ^ { - 1 }\) to \(5 \mathrm {~ms} ^ { - 1 }\) The coefficient of friction between the box and the surface is 0.4
7
  1. By using an energy method, find \(x\).
    7
  2. Describe how the model could be refined to obtain a more realistic value of \(x\) and use an energy argument to explain whether this would increase or decrease the value of \(x\).

Question 7(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Work done by tension \(= 50\cos 40° \cdot x\)B1 (AO 1.1b) Obtains correct expression for work done by tension
Resolving vertically: \(R = 8 \times 9.8 - 50\sin 40°\)M1 (AO 3.3) Finds normal reaction force using vertical component of tension
Work done against friction \(= 0.4(78.4 - 50\sin 40°)x\)A1 (AO 1.1b) Obtains correct expression for work done against friction
Work–energy principle: \(50\cos 40°\cdot x - 0.4(78.4 - 50\sin 40°)x = \frac{1}{2}\times 8\times 5^2 - \frac{1}{2}\times 8\times 2^2\)M1 (AO 3.4) Sets up model using work–energy principle including initial and final KE and work done by/against at least one force
\(84 = 19.80x\)A1 (AO 1.1b) Obtains fully correct equation using work–energy principle
\(x = 4.2\)A1 (AO 1.1b) AWRT 4.2
Question 7(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Include air resistanceM1 (AO 3.5c) States a valid refinement
Air resistance would increase \(x\) because more work would need to be done to reach the required speedE1 (AO 2.2a) Correctly deduces how this would change \(x\) with valid explanation referring to work/energy
## Question 7(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Work done by tension $= 50\cos 40° \cdot x$ | B1 (AO 1.1b) | Obtains correct expression for work done by tension |
| Resolving vertically: $R = 8 \times 9.8 - 50\sin 40°$ | M1 (AO 3.3) | Finds normal reaction force using vertical component of tension |
| Work done against friction $= 0.4(78.4 - 50\sin 40°)x$ | A1 (AO 1.1b) | Obtains correct expression for work done against friction |
| Work–energy principle: $50\cos 40°\cdot x - 0.4(78.4 - 50\sin 40°)x = \frac{1}{2}\times 8\times 5^2 - \frac{1}{2}\times 8\times 2^2$ | M1 (AO 3.4) | Sets up model using work–energy principle including initial and final KE and work done by/against at least one force |
| $84 = 19.80x$ | A1 (AO 1.1b) | Obtains fully correct equation using work–energy principle |
| $x = 4.2$ | A1 (AO 1.1b) | AWRT 4.2 |

---

## Question 7(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Include air resistance | M1 (AO 3.5c) | States a valid refinement |
| Air resistance would increase $x$ because more work would need to be done to reach the required speed | E1 (AO 2.2a) | Correctly deduces how this would change $x$ with valid explanation referring to work/energy |
7 In this question use $g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$

A box, of mass 8 kg , is on a rough horizontal surface.\\
A string attached to the box is used to pull it along the surface.\\
The string is inclined at an angle of $40 ^ { \circ }$ above the horizontal.\\
The tension in the string is 50 newtons.\\
As the box moves a distance of $x$ metres, its speed increases from $2 \mathrm {~ms} ^ { - 1 }$ to $5 \mathrm {~ms} ^ { - 1 }$\\
The coefficient of friction between the box and the surface is 0.4\\
7
\begin{enumerate}[label=(\alph*)]
\item By using an energy method, find $x$.\\

7
\item Describe how the model could be refined to obtain a more realistic value of $x$ and use an energy argument to explain whether this would increase or decrease the value of $x$.
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2020 Q7 [8]}}