A pressure-pump has a horizontal tube of cross sectional area \(10 cm^2 \) for the outflow of water at a speed of \(20 m/s\). The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is
To find the force exerted on the vertical wall that stops the water flow, we will use the concept of momentum change and Newton's second law of motion.
Given Data:
Cross-sectional area of the tube, A = 10 \, cm^2 = 10 \times 10^{-4} \, m^2.
Velocity of water, v = 20 \, m/s.
Density of water, \(\rho = 1000 \, kg/m^3\).
Step-by-Step Solution:
The volume of water flowing out per second is given by the product of the area and the velocity:
\[ Q = A \times v = (10 \times 10^{-4}) \times 20 = 2 \times 10^{-2} \, m^3/s \]
The mass flow rate, which is the mass of water flowing per second, can be calculated using the formula:
Force exerted by the water on the wall can be calculated using Newton's second law, which states that force is the rate of change of momentum. Since water is stopped completely, its final velocity is 0. Initially, it has a velocity of 20 m/s, so:
\[ F = (\text{mass flow rate}) \times \Delta v = 20 \times (20 - 0) = 400 \, N \]
Conclusion:
The force exerted on the vertical wall by the water, stopping its flow, is 400 N.
