A jet of water issues from a nozzle with a velocity of 10 m/s and it impinges normally on a fixed flat plate. The cross sectional area of the jet is 0.02 m2 and the density of water is 1000 kg/m3 . What is the force developed on the plate?

A jet of water issues from a nozzle with a velocity of 10 m/s and it impinges normally on a fixed flat plate. The cross sectional area of the jet is 0.02 m2 and the density of water is 1000 kg/m3 . What is the force developed on the plate? Correct Answer 2000 N

Concept:

Force Exerted by Jet on Moving Flat Plate Normal to Jet:

Force on the plate due to impact, F = ρa(v - u)2

where, ρ = density, a = Area of jet

v = velocity of jet, u = Plate velocity

Calculation:

Given:

Area of jet (a) = 0.02 m2

v = 10 m/s, u = 0 m/s, ρ = 1000 kg/m3

∴ Force on the plate due to impact

F = ρa(v - u)2 

F = 1000 × 0.02 × (10 - 0)2 = 2000 N

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Related Questions

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