Answer to Question #199964 in Physics for Jes

Question

The great limestone caverns were formed by dripping water. (a) If water droplets of 0.03 mL fall from a height of 5 m at a rate of 10 per minute, what is the average force exerted on the limestone floor by the droplets of water during a 1-min period?

Answer 1

Let's first find the mass of the water droplet from the definition of the density:

$$\rho=\dfrac{m}{V},$$ $$m=\rho V=1000\ \dfrac{kg}{m^3}\cdot0.03\ mL\cdot\dfrac{1.0\cdot10^{-6}\ m^3}{1\ mL}=3.0\cdot10^{-5}\ kg.$$

Then, we can find the velocity of the water droplet when it hits the limestone floor from the law of conservation of energy:

$$mgh=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot5\ m}=9.9\ \dfrac{m}{s}.$$

Let's calculate the rate of droplets per second:

$$\dfrac{N}{\Delta t}=10\ \dfrac{droplets}{1\ min}\cdot\dfrac{1\ min}{60\ s}=0.17\ \dfrac{droplets}{s}.$$

Finally we can find the average force exerted on the limestone floor by the droplets of water as follows:

$$F_{avg}=\dfrac{Nm\Delta v}{\Delta t},$$ $$F_{avg}=\dfrac{0.17\ droplets\cdot3.0\cdot10^{-5}\ kg\cdot9.9\ \dfrac{m}{s}}{1\ s}=5.05\cdot10^{-5}\ N.$$