Problem 49 - University Physics with Modern Physics

Question

A fountain sends water to a height of \(100 . \mathrm{m}\). What is the difference between the pressure of the water just before it is released upward and the atmospheric pressure?

Step-by-Step Solution

Answer

Answer: The difference between the pressure of the water just before it is released and the atmospheric pressure is \(9.81 \times 10^5 \, \mathrm{Pa}\).

Step 1: Identify the known values and unknowns

We are given the height of the water at the top of the fountain (\(h = 100\) meters). We need to find the difference between the pressure at the base of the fountain and atmospheric pressure.

Step 2: Find the potential energy at the top

Using the given height, we can find the potential energy at the top of the fountain. The potential energy (\(U\)) is given by the formula: \( U = mgh \) where \(m\) is the mass of the water, \(g\) is the acceleration due to gravity (which is approximately \(9.81 \, \frac{m}{s^2}\)), and \(h\) is the height of the water (\(100 \, m\)).

Step 3: Find the kinetic energy at the top

Since the water reaches its maximum height, its vertical velocity at the top is \(0\). Therefore, its kinetic energy (\(K\)) at the top is also \(0\).

Step 4: Apply Bernoulli's equation

Bernoulli's equation is given by: \(P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2\) At the top of the fountain (\(P_2\), \(v_2\) and \(h_2\)), the pressure is equal to atmospheric pressure, the velocity is zero, and the height is \(100 \,m\). At the base of the fountain, we need to find the pressure \(P_1\), the velocity \(v_1\) and the height \(h_1\) (which is \(0\)). Substitute the known values into Bernoulli's equation: \(P_1 + \frac{1}{2} \rho v_1^2 = P_{atm} + \rho g h\) Since the kinetic energy at the top is \(0\), the equation becomes: \(P_1 = P_{atm} + \rho g h\)

Step 5: Calculate the difference in pressure

To find the difference between the pressure at the base of the fountain (just before the water is released) and the atmospheric pressure, we can subtract \(P_{atm}\) from both sides of the equation: \(\Delta P = P_1 - P_{atm} = \rho g h\) We don't have the density of water in the problem, but it is approximately \(\rho = 1000 \, \frac{kg}{m^3}\). Plug in the values for density, gravity, and height: \(\Delta P = 1000 \, \frac{kg}{m^3} \cdot 9.81 \, \frac{m}{s^2} \cdot 100 \, m\) So, \(\Delta P = 9.81 \times 10^5 \, \mathrm{Pa}\) The difference between the pressure of the water just before it is released and the atmospheric pressure is \(9.81 \times 10^5 \, \mathrm{Pa}\).