Answer to Question #160003 in Mechanics | Relativity for Emmanuel

Question

A hose directs a horizontal jet of water, moving with a velocity of 20m/s on to a vertical wall. The cross sectional area of the jet is 5 * 10^-4. If the density of water is 1000kg/m^3, calculate the force on a wall assuming the water is brought to rest there.

Answer 1

Let's first find the volume flow rate of water that flows through a hose per second:

$$V=Av=5\cdot10^{-4}\ m^2\cdot20\ \dfrac{m}{s}=0.01\ \dfrac{m^3}{s}.$$

Then, we can find the mass flow rate or the mass that hits the wall each second:

$$m=\rho V=1000\ \dfrac{kg}{m^3}\cdot0.01\ \dfrac{m^3}{s}=10\ \dfrac{kg}{s}.$$

Finally, we can find the force acting on a wall:

$$F\Delta t=m\Delta v,$$ $$F=\dfrac{m(v_f-v_i)}{\Delta t}=\dfrac{10\ kg\cdot(0-20\ \dfrac{m}{s})}{1\ s}=-200\ N.$$

The magnitude of the force on the wall is 200 N. The sign minus means that the force on the wall directed away from the wall.