AQA AS Level Physics Paper 2, November 2021: Question 4
9 marks · Hard difficulty · Extended Answer
Analyze forces, acceleration, and power requirements for concrete blocks being raised on an inclined conveyor belt driven by an electric motor.
Practise this questionQuestion
Question text
04 Figure 10 shows a conveyor used to raise concrete blocks on a building site.
The blocks do not slip on the belt at any time.
Figure 10
Figure 11 shows an enlarged view of one block on the belt. The belt is inclined at 23°
to the horizontal. The mass of the block is 19 kg.
Figure 11
The belt exerts a frictional force F on the block when the block is at rest.
04.1 Draw an arrow on Figure 11 to show the line of action of F.
[1 mark]
04.2 Show that the magnitude of F is approximately 70 N.
[1 mark]
04.3 The belt is driven by an electric motor. When the motor is switched on, the belt and
the block accelerate uniformly from rest to a speed of 0.32 m s−1 in a time of 0.50 s.
Calculate the magnitude of the frictional force of the belt on the block during this
acceleration.
[3 marks]
14 N
frictional force =
04.4 The motor is connected to a 110 V dc supply that has negligible internal resistance.
The maximum operating current in the motor is 5.0 A.
The efficiency of the motor and drive system of the conveyor is 28%. The belt travels
at 0.32 m s−1 and is 8.0 m long.
Deduce the maximum number of blocks that can be moved on the belt at one time.
[4 marks]
maximum number of blocks =
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
04.1 arrow between block and belt pointing upwards along the belt 1 AO1.1-a
04.2 (F = ) 19gsin23° to give 72.8 (N) Allow 2 sf answer. 1 AO2.1-f
04.3 ∆(mv) Allow for MP1 use of appropriate kinematic 3 AO2.1-b
uses F =
∆t equation for a AND use of F=ma
F = 12 (N)
their 04.2 + 12 (N) Expect 82 or 85 (N)
– SICS – – JUNE 2021
04.4 uses V and I to get total input power or energy Pinput of motor = 110 × 5.0 = 550 W 4 AO3.1-a
8.0
Einput = 550 × 0.32 = 13 750 J
uses efficiency equation Puseful to belt = 550 × 0.28 = 150 W
8.0
Euseful = 3850 J, from 154 × 0.32 , or 13 750 × 0.28
determines power or energy to move one block Pblock = 22 or 23 W
Eblock = 560 or 580 J
divides (total) useful power or energy by individual power or Allow ecf for MP4 only for their 04.3
energy to give answer of 6 blocks
Total 9
How to answer it
Forces, Dynamics and Motor Power Study Guide
What this question tests
This question assesses core mechanics and electrical power concepts at AS Level. Key topics include resolving forces on inclined planes, applying Newton's laws of motion (F = ma and rate of change of momentum), combining static and dynamic friction components, and calculating electrical power input versus useful mechanical output using efficiency.
Direction of the Frictional Force
Drawing the line of action of friction on an incline
✅ Correct Answer
An arrow drawn clearly between the block and the belt, pointing upwards and parallel to the belt.
💡 Key Knowledge
Friction always acts parallel to the surfaces in contact and opposes relative motion (or the tendency of motion). Since gravity pulls the block down the slope, the frictional force must act up the slope to keep it at rest.
Static Equilibrium Calculation
Showing the magnitude of the frictional force at rest
📐 Calculation Steps
- Identify the component of weight acting down the slope: W_parallel = mg sin(theta)
- Substitute values: 19 × 9.81 × sin(23°) or using g = 9.8
- Calculate result: 72.8 N (matches approximately 70 N as stated in the prompt).
🧠 Exam Technique
When a question says "Show that...", your calculated value must be slightly more precise than the stated figure (e.g., show 72.8 N to justify ≈ 70 N ). Never just write 70 N without showing working.
Dynamic Acceleration and Total Force
Calculating friction during belt acceleration
📐 Step-by-Step Calculation
- Find acceleration (a):
a = (v - u) / t = (0.32 - 0) / 0.50 = 0.64 m s⁻² - Calculate force needed for acceleration (F = ma):
F_accel = 19 × 0.64 = 12.16 N - Combine with component of weight:
Total F = F_static + F_accel = 72.8 + 12.16 = 84.96 N (Expect 82 N or 85 N depending on rounded intermediate values).
❌ Common Errors
Students often forget to add the static component from 04.2, calculating only the force required to accelerate the mass ( ma ) and missing the component of weight acting down the slope.
System Efficiency and Capacity
Deducing the maximum number of blocks
💡 Key Knowledge & Power Relations
Power input is calculated via P = VI . Useful power transferred to the belt takes efficiency into account ( P_useful = P_input × efficiency ). Power required per block can be found using P = F × v or via energy over time.
📐 Step-by-Step Solution
- Input Power: P_in = 110 V × 5.0 A = 550 W
- Useful Power to Belt: P_useful = 550 × 0.28 = 154 W
- Power per Block:
P_block = F × v = 84.96 N × 0.32 m s⁻¹ ≈ 27.2 W (or using energy methods over the 8.0 m length giving E_block ≈ 560 J and P_block ≈ 22-23 W based on mark scheme values). - Maximum Blocks:
Number = P_useful / P_block = 154 / 27.2 ≈ 5.66 → Round down to whole blocks: 6 blocks.
🧠 Examiner Insight & Top-Level Responses
Top-level responses systematically structure their units, converting cleanly between power and total energy over the belt length. Notice that rounding up to 7 blocks would exceed the motor's capability, so students must understand physical constraints requiring rounding down.
Topics
Physics · 3.4 Mechanics and materials · 3.5 Electricity
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.