Question

An engine pumps water continuously through a hose pipe. If the water leaves the pipe with velocity $v$ and $m$ is the mass per unit length of the water in the pipe, then the rate at which kinetic energy is imparted to the water is:

• A. $\frac{1}{2} m v^3$
• B. $\frac{1}{2} m v^2$
• C. $m v^3$
• D. $\frac{3}{2} m v^2$

Hint:

Power can be represented as $F \cdot v$. The thrust force on a pipe discharging fluid is $F = v \frac{dM}{dt} = m v^2$. Thus, Power $= F \cdot v = (m v^2) v = m v^3$. However, only half of this work goes into the water's kinetic energy, yielding $\frac{1}{2} m v^3$.

Answer 1

The Correct Option is A

Step 1: Understand what is asked.
Water is pumped through a pipe and leaves with speed $v$. We want the rate at which kinetic energy is given to the water. Rate of energy is just power.

Step 2: Recall kinetic energy and power.
Kinetic energy of mass $M$ moving at speed $v$ is $\dfrac{1}{2}Mv^2$. Power is how fast this energy is added, so \[ P = \frac{dK}{dt} = \frac{1}{2}\left(\frac{dM}{dt}\right)v^2 \] The speed $v$ stays the same, so only the mass keeps flowing.

Step 3: Find the mass flow rate.
Here $m$ is the mass per unit length of water in the pipe. In a small time, the water moves a small length. The mass passing per second is \[ \frac{dM}{dt} = m\frac{dx}{dt} = m v \] because $\dfrac{dx}{dt}$ is just the speed $v$.

Step 4: Put the mass flow into the power formula.
\[ P = \frac{1}{2}(mv)\,v^2 \]

Step 5: Simplify.
Multiply the $v$ terms together. \[ P = \frac{1}{2}m v^3 \]

Step 6: Read the meaning.
The power grows with the cube of speed. Faster water needs much more power. So the rate of kinetic energy given to the water is \[ \boxed{P = \frac{1}{2}m v^3} \]