Question

a stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. the incident water stream has a velocity of +16.0 m/s, while the exiting water stream has a velocity of -16.0 m/s. the mass of water per second that strikes the blade is 20.0 kg/s. find the magnitude of the average force exerted on the water by the blade.

Explanation:

Step1: Use the impulse - momentum theorem

The impulse - momentum theorem is \(F\Delta t=\Delta p = m(v_f - v_i)\). We can rewrite it for force per unit time. Since \(F=\frac{\Delta p}{\Delta t}\) and \(\frac{m}{\Delta t}\) is given as \(20.0\space kg/s\), \(v_i = + 16.0\space m/s\), \(v_f=-16.0\space m/s\).

Step2: Calculate the force

Substitute the values into the formula \(F=\frac{m}{\Delta t}(v_f - v_i)\). So \(F=(20.0\space kg/s)\times(-16.0 - 16.0)\space m/s\).

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The magnitude of the force is \(|F|\).

Answer:

\(640\space N\)