Question

A jet of water issues from a nozzle with a velocity of 20 m/s and it impinges normally on a flat plate moving away from it at 10 m/s. If the cross-sectional area of the jet is 0.02 m2 and the density of water is taken as 1000 kg/m3 , then the force developed on the plate will be

Answer 1

The force exerted by the water jet on the plate will be 2000 N.

Explanation :

Force exerted on the moving plate can be calculated using the relation,

F = ρA(V−U)²

where,

ρ = density of the fluid

A=area of cross section of the jet

V= velocity of water jet

U=velocity of moving plate,

U - V will be the relative velocity of water jet

Putting all the values in the equation;

F  = 1000 x 0.02 x(20-10)²

= 1000 x 0.02 x 100

= 2000 N

Hence the force exerted by the water jet on the plate will be 2000 N.