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)15 m/s
B)20 m/s
C)25 m/s
D)30 m/s

Question

Quizplus · Accepted Answer

The force required to knock down the door is 1000 N. This force is equal to the rate of change of momentum (Newton's second law) of the water stream hitting the door.

Rate of change of momentum = force = 1000 N

Let the velocity of the water stream be v. Then, the rate of flow of water (mass per second) is given as 40 kg/s.

Rate of flow of water x velocity of water = rate of change of momentum

40 kg/s x v = 1000 N

v = 1000/40 = 25 m/s

Therefore, the minimum velocity of the water stream needed to knock down the door is 25 m/s.

A Fire Hose Is Turned on the Door of a Burning