A hose ejects water at a speed of 2m/s through a hole of area 0.01m2.if the water strikes a wall normally,calculate the force on the wall in Newtons,assuming the velocity of water nomal to wall is zero after collision.
Force = (mass flow rate)(water velocity)
mass flow rate = (water density)*(water velocity)*(hole area) = 1000*2*0.02 = 40 kg/s
Solve for Force
40 N
To calculate the force exerted on the wall by the water, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a):
F = m * a
First, we need to find the mass of the water that is ejected through the hose per second. We can do this by multiplying the water's density (ρ) by its volume (V).
The density of water is approximately 1000 kg/m^3.
The volume of water per second is equal to the water's speed (v) multiplied by the area of the hole (A).
v = 2 m/s
A = 0.01 m^2
V = v * A
Next, we need to calculate the mass (m) by multiplying the density (ρ) by the volume (V):
m = ρ * V
Now, we can calculate the acceleration (a), which is the change in velocity per second. Since the velocity of the water after hitting the wall is zero, the change in velocity is equal to the initial velocity (v).
a = v / t = v / 1 s = v.
Finally, we can substitute the values into Newton's second law to find the force (F):
F = m * a
Let's calculate it step by step:
1. Calculate the volume of water per second:
V = v * A
V = 2 m/s * 0.01 m^2
V = 0.02 m^3/s
2. Calculate the mass of the water:
m = ρ * V
m = 1000 kg/m^3 * 0.02 m^3/s
m = 20 kg/s
3. Calculate the force exerted on the wall:
F = m * a
F = 20 kg/s * 2 m/s
F = 40 N
Therefore, the force exerted on the wall by the water is 40 Newtons.