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Consult Interactive Solution \(\underline{7} .9\) at for a review of problem- solving skills that are involved in this problem. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of \(+16.0 \mathrm{~m} / \mathrm{s},\) while the exiting water stream has a velocity of \(-16.0 \mathrm{~m} / \mathrm{s}\). The mass of water per second that strikes the blade is \(30.0 \mathrm{~kg} / \mathrm{s}\). Find the magnitude of the average force exerted on the water by the blade.

Short Answer

Expert verified
The magnitude of the force is 960 N.

Step by step solution

01

Identify Known Quantities

The problem provides certain quantities: \( v_i = +16.0 \ \mathrm{m/s} \), which is the initial water velocity, \( v_f = -16.0 \ \mathrm{m/s} \) for the final velocity, and \( \dot{m} = 30.0 \ \mathrm{kg/s} \) for the mass flow rate (mass of water per second).
02

Determine the Change in Velocity

Calculate the change in velocity (\( \Delta v \)) of the water stream as it strikes and leaves the blade: \( \Delta v = v_f - v_i = -16.0 - (+16.0) = -32.0 \ \mathrm{m/s} \).
03

Calculate the Change in Momentum

Momentum change per second (force) can be calculated using the product of mass flow rate and change in velocity:\( \Delta p = \dot{m} \times \Delta v = 30.0 \ \mathrm{kg/s} \times (-32.0 \ \mathrm{m/s}) = -960 \ \mathrm{kg \cdot m/s^2} \).
04

Calculate the Magnitude of Force

The magnitude of force exerted on the water by the blade is the absolute value of the change in momentum per second:\[ |F| = |-960 \ \mathrm{N}| = 960 \ \mathrm{N} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum
Momentum is an essential concept in physics, describing the quantity of motion an object possesses. It is the product of an object's mass and velocity. In mathematical terms, it's expressed as \( p = m imes v \), where \( p \) is the momentum, \( m \) is mass, and \( v \) is velocity. This concept is crucial when analyzing the motion of objects and understanding how forces interact with these objects.
For systems like the water hitting a turbine blade, momentum conservation principles are often applied. When the water strikes the blade, it undergoes a change in its momentum due to the force applied by and on the blade.
  • Initial momentum \( = m imes v_i \)
  • Final momentum \( = m imes v_f \)
  • Change in momentum \( = m \times (v_f - v_i) \)
This change helps us understand how forces affect moving objects, such as in power generation or braking systems.
Force Calculation
Force is a fascinating quantity in physics that describes the interaction between bodies resulting in motion or a change of state. When calculating force, especially in systems where mass flow and velocity change, such as fluid dynamics, Newton's second law is key. This law describes how force equals the rate of change in momentum.
To calculate force exerted by the turbine blade on the water, we need the mass flow rate (mass per second) and the change in velocity. If we have the mass flow rate \( \dot{m} \) and the velocity change \( \Delta v \), the force \( F \) can be calculated as:\[ F = \dot{m} \times \Delta v \]This equation links force to the inherent changes in the momentum of the water as it interacts with the turbine. It is often employed to measure and control forces in hydraulic systems and engines. The \'-\' in the calculation signifies the direction of force and is crucial in understanding how blades transfer energy.
Velocity Change
Understanding velocity change is key to solving motion-related problems. Velocity describes an object's speed in a given direction, and a change in velocity indicates that a force or interaction has occurred.
In this exercise, the velocity change \( \Delta v \) is the difference between the initial and final velocity of the water:
\[ \Delta v = v_f - v_i \]
Here, this value is \( -32.0 \, \mathrm{m/s} \), which tells us that the water has reversed its direction after hitting the turbine blade. Such a significant change signifies a substantial external influence, usually exerted as force to alter the motion. This concept is pivotal in mechanics, rocket propulsion, and any scenario where speed and direction of objects are manipulated.
Mass Flow Rate
Mass flow rate is an important parameter in fluid dynamics, describing how much mass flows through a system per unit time. It is usually denoted by \( \dot{m} \) and measured in \( \mathrm{kg/s} \). Understanding this concept is key in processes such as chemical reactions, HVAC systems, and fluid transport.
In the context of the turbine problem, the mass flow rate of water is given as \( 30.0 \, \mathrm{kg/s} \). This informs us about how much water, by mass, is impacting the turbine blade every second. When analyzing force, the mass flow rate serves as a critical component:
  • It's used to calculate the change in momentum.
  • It helps determine how force is distributed over time.
The mass flow rate ensures that systems, whether in engines, power plants, or even in natural flows like rivers, are sized and managed to accommodate this specific flow of matter.

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Most popular questions from this chapter

A wagon is coasting at a speed \(v_{\mathrm{A}}\) along a straight and level road. When ten percent of the wagon's mass is thrown off the wagon, parallel to the ground and in the forward direction, the wagon is brought to a halt. If the direction in which this mass is thrown is exactly reversed, but the speed of this mass relative to the wagon remains the same, the wagon accelerates to a new speed \(v_{\mathrm{B}}\). Calculate the ratio \(v_{\mathrm{B}} / v_{\mathrm{A}}\).

Part \(a\) of the drawing shows a bullet approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part \(b\). Note that both blocks are moving after the collision with the bullet. (a) Can the conservation of linear momentum be applied to this three-object system, even though the second collision occurs a bit later than the first one? Justify your answer. Neglect any mass removed from the first block by the bullet. (b) Is the total kinetic energy of this three-body system conserved? If not, would the total kinetic energy after the collisions be greater than or smaller than that before the collisions? Justify your answer.

A \(0.500-\mathrm{kg}\) ball is dropped from rest at a point \(1.20 \mathrm{~m}\) above the floor. The ball rebounds straight upward to a height of \(0.700 \mathrm{~m}\). What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

A car (mass \(=1100 \mathrm{~kg}\) ) is traveling at \(32 \mathrm{~m} / \mathrm{s}\) when it collides head-on with a sport utility vehicle (mass \(=2500 \mathrm{~kg}\) ) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?

A 50.0 -kg skater is traveling due east at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). A \(70.0-\mathrm{kg}\) skater is moving due south at a speed of \(7.00 \mathrm{~m} / \mathrm{s}\). They collide and hold on to each other after the collision, managing to move off at an angle \(\theta\) south of east, with a speed of \(v_{\mathrm{f}}\). Find (a) the angle \(\theta\) and (b) the speed \(v_{\mathrm{f}}\), assuming that friction can be ignored.

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