Question

In a waterfall of height 100 m, 2000 kg of water falls per second. Calculate the potential energy of water at its initial position. Determine the final kinetic energy of water on the ground.

Answer 1

Given Data:

Height of the waterfall (h) = 100 m

Mass of water falling per second (m) = 2000 kg

Acceleration due to gravity (g) = 10 m/s²

1. Potential energy (PE) is calculated using the formula:

PE = m × g × h 

Since 2000 kg of water falls every second, the potential energy possessed by the water per second is:

PE = 2000 kg × 10 m/s² × 100 m

= 2,000,000 J/s

= 2 × 10⁶ J/s

2. Just before striking the ground, all the potential energy converts into kinetic energy. However, once the falling water hits the ground, it immediately comes to a complete rest.

As a result, its final velocity becomes zero (v = 0), causing all its kinetic energy to be converted into heat and sound energy.

Therefore, the final kinetic energy of the water after it has settled on the ground is:

KE = 0