Answer to Question #170609 in Mechanics | Relativity for Mae

Question

A fire hose ejects water at the rate of 60 kg of water per second at a velocity of 40 m/s. With what force must a fireman hold on the hose to keep it stationary?

Answer 1

$$F\Delta t=\Delta p=m\Delta v,$$ $$F\Delta t=mv_f-mv_i=-mv_i,$$ $$F=\dfrac{-mv_i}{\Delta t}=\dfrac{-60\ kg\cdot40\ \dfrac{m}{s}}{1\ s}=-2400\ N.$$

The magnitude of the force is 2400 N. The sign minus means that the force on the hose is directed backward. So, the fireman must hold on the hose with the force of 2400 N to keep it stationary.