AP · AP 1 · extended response
A hydraulic lift below the reservoir surface
AP Physics 1: Algebra-Based · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Hydrostatic gauge-pressure contribution below the small piston
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Additional pressure / Pa
Horizontal axis (x): Depth below small piston / m
Read the exact data as a table
| Depth below small piston / m | Additional pressure / Pa |
|---|---|
| 0 | 0 |
| 0.25 | 2000 |
| 0.5 | 4000 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Choose your AP course · School year 2026–27 · May 2027 exam
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Choose reviewed AP practice →An ideal hydraulic system contains oil of density 800 kg m⁻³. Its small piston has area 2.0 cm² and its large piston has area 40 cm². The large piston is 0.50 m below the small piston. A 60 N force is applied downward to the small piston; atmospheric pressure acts on both external faces. Use g = 10 N kg⁻¹ and neglect piston weights and friction. Supplied data (Depth below small piston / m; Additional pressure / Pa): (0, 0), (0.25, 2000), (0.5, 4000).
- (a)
Find the gauge pressure immediately below each piston and the greatest downward load the large piston can balance.
3 marks - (b)
For a short motion, the small piston moves down 0.10 m. Find the large-piston rise and explain why force amplification does not create energy.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find the gauge pressure immediately below each piston and the greatest downward load the large piston can balance.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Convert 2.0 cm² to 2.0×10⁻⁴ m². Applied gauge pressure is F/A = 60/(2.0×10⁻⁴) = 300000 Pa.
- 2
The lower piston has an additional ρgh = 800(10)(0.50) = 4000 Pa, giving 304000 Pa gauge.
- 3
Its area is 40 cm² = 0.0040 m², so the balanced load is pA = 304000(0.0040) = 1216 N. Atmospheric contributions cancel across each piston.
(b)
For a short motion, the small piston moves down 0.10 m. Find the large-piston rise and explain why force amplification does not create energy.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Incompressibility gives equal displaced volumes: As xs = AL xL, so xL = (2/40)(0.10) = 0.0050 m.
- 2
The large force acts through a smaller displacement. For equal piston heights the ideal input and output work are equal.
- 3
Here the initial pressure difference also involves gravitational potential energy of the oil. For a finite stroke the height difference changes; a full energy audit must include the oil and the varying pressure, rather than multiplying the initial force by any arbitrary stroke.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Convert 2.0 cm² to 2.0×10⁻⁴ m². Applied gauge pressure is F/A = 60/(2.0×10⁻⁴) = 300000 Pa.
- The lower piston has an additional ρgh = 800(10)(0.50) = 4000 Pa, giving 304000 Pa gauge.
- Its area is 40 cm² = 0.0040 m², so the balanced load is pA = 304000(0.0040) = 1216 N. Atmospheric contributions cancel across each piston.
- Incompressibility gives equal displaced volumes: As xs = AL xL, so xL = (2/40)(0.10) = 0.0050 m.
- The large force acts through a smaller displacement. For equal piston heights the ideal input and output work are equal.
- Here the initial pressure difference also involves gravitational potential energy of the oil. For a finite stroke the height difference changes; a full energy audit must include the oil and the varying pressure, rather than multiplying the initial force by any arbitrary stroke.
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