Question

A hydraulic cylinder has piston (cross-sectional area 25 cm²) and fluid pressure
of 2 MPa. If piston moves 0.25 m. how much work is done?
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Answer 1

Answer:

The work is a force with a displacement and force is constant: F = PA W = ∫ F dx = ∫ PA dx = PA ∆x = 2000 kPa × 25 × 10-4 m2 × 0.25 m = 1.25 kJ Units: kPa m2 m = kN m-2 m2 m = kN m = kJRead more on Sarthaks.com - https://www.sarthaks.com/453677/hydraulic-cylinder-has-piston-cross-sectional-area-and-fluid-pressure-mpa-the-piston-moved

Explanation:

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Answer 2

Answer:

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