A fire hose is turned on the door of a burning building in order to knock the door down. This requires a force of 1000 N. If the hose delivers 40 kg of water per second, what is the minimum
Velocity of the stream needed to knock down the door, assuming the water doesn't bounce back?
A) 20 m/s
B) 30 m/s
C) 25 m/s
D) 15 m/s

Question

Quizplus · Accepted Answer

To knock down the door, the force exerted by the water must equal the required force of 1000 N. The force exerted by the water can be calculated using the formula $F = \Delta p / \Delta t$, where $F$ is the force, $\Delta p$ is the change in momentum, and $\Delta t$ is the change in time. Since the water doesn't bounce back, its final velocity is 0, and the change in momentum is given by $\Delta p = m \Delta v$, where $m$ is the mass of water per second (40 kg/s) and $\Delta v$ is the change in velocity, which is equal to the velocity of the water stream because it starts from rest. Therefore, $F = m v / \Delta t$. Given that $F = 1000 N$ and $m/\Delta t = 40 kg/s$, we can solve for $v$: $1000 = 40v$, which gives $v = 25 m/s$.

A Fire Hose Is Turned on the Door of a Burning