O-LevelPhysicsForces5054/22 · May/June 2014 · Q26 Marks
Fig. 2.1 shows part of a hydraulic jack used to lift the front of a car. The operator pulls the handle and causes a force of 50N to act on the small piston. The force exerted by the oil on the large piston increases by F. The large piston moves and rotates the arm about the pivot. This raises the front of the car. The cross-sectional area of the small piston is 1.5cm². The cross-sectional area of the large piston is 5.0cm². [Figure 2.1]
- (a)(i)Calculate the pressure in the oil caused by the force on the small piston,[2]
- (a)(ii)Calculate the value of F.[1]
- (b)Explain why the large piston moves through a shorter distance than the small piston.[1]
- (c)The efficiency of the jack is 75%. Explain what is meant by efficiency.[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.
- [C1] (a) (i) (P=) F/A algebraic in any form or numerical
- [A1] 33 N/cm² or 3.3 × 105 N/m²
- [B1] (ii) 170 N or 167 N or 166.7 N or (i) × 5 with unit
- [B1] (b) volume (of oil) remains the same
- or oil passes from small(er) to large(r) area
- or work = Fd and force large so distance small
- [C1] (c) output ÷ input or fraction or percentage of work mentioned
- [A1] complete definition, e.g. useful work obtained ÷ (total) work put in
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