AQA Further Paper 3 Mechanics 2021 June — Question 8 11 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2021
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeLifting objects vertically
DifficultyChallenging +1.2 This is a Further Maths mechanics question requiring integration of a variable force and application of work-energy principle across multiple parts. Part (a) is straightforward integration (showing a given result), parts (b-c) require systematic application of work-energy theorem with careful bookkeeping of limits, and part (d) is a simple explanation. While it involves several steps and the variable force adds complexity beyond standard constant-force problems, the techniques are standard Further Maths fare with no novel insights required—making it moderately above average difficulty but not exceptional.
Spec6.02b Calculate work: constant force, resolved component6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle
In this question use \(g = 9.8 \text{ m s}^{-2}\) A lift is used to raise a crate of mass 250 kg The lift exerts an upward force of magnitude \(P\) newtons on the crate. When the crate is at a height of \(x\) metres above its initial position $$P = k(x + 1)(12 - x) + 2450$$ where \(k\) is a constant. The crate is initially at rest, at the point where \(x = 0\)
  1. Show that the work done by the upward force as the crate rises to a height of 12 metres is given by $$29400 + 360k$$ [3 marks]
  2. The speed of the crate is \(3 \text{ m s}^{-1}\) when it has risen to a height of 12 metres. Find the speed of the crate when it has risen to a height of 15 metres. [5 marks]
  3. Find the height of the crate when its speed becomes zero. [2 marks]
  4. Air resistance has been ignored. Explain why this is reasonable in this context. [1 mark]
Question 8:
AnswerMarks
8(a)Recognises that WD is the integral
of force with respect to x
AnswerMarks Guidance
Condone missing dx3.3 M1
W D =  k ( x + 1 ) (1 2 − x ) + 2 4 5 0 d x
0
1 2 1 2
= k  ( x + 1 ) (1 2 − x ) d x +  2 4 5 0 d x
0 0
= 3 6 0 k + 2 9 4 0 0
Obtains correct integral with correct
AnswerMarks Guidance
limits1.1b A1
Obtains correct expression for the
work done from a rigorous
AnswerMarks Guidance
argument2.1 R1
Total3
QMarking Instructions AO
AnswerMarks Guidance
8(b)Forms an energy equation to find k 3.3
2 9 4 0 0 + 3 6 0 k = 2 5 0  9 8.  1 2 +  2 5 0  3 2
2
3 6 0 k = 1 1 2 5
2 5
k =
8
2 5 5 8 5
W o rk D o n e = 3 6 7 5 0 + 
8 2
2 5 5 8 5 1
3 6 7 5 0 +  = 2 5 0 9 8 1 5  .  + 2  5 0  v 2
8 2 2
2 2 5 5 8 5  
v =
1 6 2 5 0 
= 2 7 0 4 2 7 m . = . 1− s to 2 s f
AnswerMarks Guidance
Obtains the correct value for k1.1b A1
Finds the work done when the
crate has risen from 0 to15 metres
AnswerMarks Guidance
or from 12 to 15 metres, FT their k1.1a M1
Forms an energy equation to find
AnswerMarks Guidance
the speed at 15 metres3.4 M1
Finds the correct speed and gives
the answer to 2 sig fig, with correct
AnswerMarks Guidance
units3.2a A1
Total5
QMarking Instructions AO
AnswerMarks Guidance
8(c)Forms an equation to find x (PI) 3.4
12kx+ − =0
2 3
 11x x2
kx 12+ − =0
 
 2 3 
x=0, 18.45, −1.95
x=18 metres to 2 sf
Solves the equation and deduces
correct value of x, condone missing
units.
AnswerMarks Guidance
AWRT 182.2a A1
Total2
QMarking Instructions AO
AnswerMarks Guidance
8(d)Gives valid explanation with
reference to low speeds3.5a E1
the crate is very low
AnswerMarks Guidance
Total1
Question total11
QMarking Instructions AO 
Question 8:
--- 8(a) ---
8(a) | Recognises that WD is the integral
of force with respect to x
Condone missing dx | 3.3 | M1 | 1 2
W D =  k ( x + 1 ) (1 2 − x ) + 2 4 5 0 d x
0
1 2 1 2
= k  ( x + 1 ) (1 2 − x ) d x +  2 4 5 0 d x
0 0
= 3 6 0 k + 2 9 4 0 0
Obtains correct integral with correct
limits | 1.1b | A1
Obtains correct expression for the
work done from a rigorous
argument | 2.1 | R1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
--- 8(b) ---
8(b) | Forms an energy equation to find k | 3.3 | M1 | 1
2 9 4 0 0 + 3 6 0 k = 2 5 0  9 8.  1 2 +  2 5 0  3 2
2
3 6 0 k = 1 1 2 5
2 5
k =
8
2 5 5 8 5
W o rk D o n e = 3 6 7 5 0 + 
8 2
2 5 5 8 5 1
3 6 7 5 0 +  = 2 5 0 9 8 1 5  .  + 2  5 0  v 2
8 2 2
2 2 5 5 8 5  
v =
1 6 2 5 0 
= 2 7 0 4 2 7 m . = . 1− s to 2 s f
Obtains the correct value for k | 1.1b | A1
Finds the work done when the
crate has risen from 0 to15 metres
or from 12 to 15 metres, FT their k | 1.1a | M1
Forms an energy equation to find
the speed at 15 metres | 3.4 | M1
Finds the correct speed and gives
the answer to 2 sig fig, with correct
units | 3.2a | A1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
--- 8(c) ---
8(c) | Forms an equation to find x (PI) | 3.4 | M1 | 11kx2 kx3
12kx+ − =0
2 3
 11x x2
kx 12+ − =0
 
 2 3 
x=0, 18.45, −1.95
x=18 metres to 2 sf
Solves the equation and deduces
correct value of x, condone missing
units.
AWRT 18 | 2.2a | A1
Total | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 8(d) ---
8(d) | Gives valid explanation with
reference to low speeds | 3.5a | E1 | Reasonable because the speed of
the crate is very low
Total | 1
Question total | 11
Q | Marking Instructions | AO | Marks | Typical Solution
In this question use $g = 9.8 \text{ m s}^{-2}$

A lift is used to raise a crate of mass 250 kg

The lift exerts an upward force of magnitude $P$ newtons on the crate.

When the crate is at a height of $x$ metres above its initial position

$$P = k(x + 1)(12 - x) + 2450$$

where $k$ is a constant.

The crate is initially at rest, at the point where $x = 0$

\begin{enumerate}[label=(\alph*)]
\item Show that the work done by the upward force as the crate rises to a height of 12 metres is given by

$$29400 + 360k$$
[3 marks]

\item The speed of the crate is $3 \text{ m s}^{-1}$ when it has risen to a height of 12 metres.

Find the speed of the crate when it has risen to a height of 15 metres.
[5 marks]

\item Find the height of the crate when its speed becomes zero.
[2 marks]

\item Air resistance has been ignored.

Explain why this is reasonable in this context.
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2021 Q8 [11]}}
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