Water is pumped from a stone basin up through a pipe 1 meter high. At that height, the water flows out through a tap and falls down through the air to the basin below, where the cycle begins again. What is the gravitational potential energy of 2.5 kilograms of water at the top of the pipe? How fast is the falling water moving by the time it reaches the basin?
To find the gravitational potential energy of the water at the top of the pipe, we can use the formula:
Gravitational Potential Energy = mass * gravitational acceleration * height
Given:
Mass of water (m) = 2.5 kg
Height (h) = 1 meter
Gravitational acceleration (g) on Earth is approximately 9.8 m/s^2
Substituting these values into the formula:
Gravitational Potential Energy = 2.5 kg * 9.8 m/s^2 * 1 m
Gravitational Potential Energy = 24.5 Joules
Thus, the gravitational potential energy of 2.5 kilograms of water at the top of the pipe is 24.5 Joules.
Now let's calculate the speed of the falling water when it reaches the basin. Since we are considering the motion of a falling object, we can use the principle of conservation of energy. At the top of the pipe, all of the gravitational potential energy is converted into kinetic energy (KE) as the water falls. The formula for kinetic energy is:
Kinetic Energy = (1/2) * mass * velocity^2
Setting the gravitational potential energy equal to the kinetic energy:
Gravitational Potential Energy = Kinetic Energy
Using the values we know from the previous calculation:
24.5 J = (1/2) * 2.5 kg * velocity^2
Simplifying the equation:
49 = 2.5 * velocity^2
velocity^2 = 49 / 2.5
velocity^2 = 19.6
velocity ≈ √19.6
velocity ≈ 4.43 m/s
Therefore, the falling water will be moving at approximately 4.43 m/s when it reaches the basin.