An engine pumps water continously through a hole. The speed with which water pases through the hole nozzle is v and k is the mass per unit length of the water jet as it leaves the nozzle. Find the rate at which kinetic energy is being imparted to the water.

Options

A. `1/2 kv^2`
B. `1/2 kv^3`
C. `(v^2)/(2k)`
D. `(v^3)/(2k)`

Explanation

To find the rate at which kinetic energy is being imparted to the water, we can follow these steps:

### Step 1: Understand the relationship between mass and length
We are given that \( k \) is the mass per unit length of the water jet. This can be expressed mathematically as:
\[ k = \frac{dm}{dx} \]
where \( dm \) is the mass of the water and \( dx \) is the length of the water jet.

### Step 2: Define the kinetic energy
The kinetic energy (\( KE \)) of a mass \( m \) moving with a velocity \( v \) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]

### Step 3: Express the rate of change of kinetic energy
To find the rate at which kinetic energy is being imparted to the water, we need to calculate the time derivative of kinetic energy:
\[ \frac{d(KE)}{dt} = \frac{d}{dt}\left(\frac{1}{2} mv^2\right) \]

### Step 4: Apply the chain rule
Using the chain rule, we can express this as:
\[ \frac{d(KE)}{dt} = \frac{1}{2} \cdot v^2 \cdot \frac{dm}{dt} \]

### Step 5: Relate mass flow rate to velocity
The mass flow rate (\( \frac{dm}{dt} \)) can be expressed in terms of the mass per unit length and the velocity of the water:
\[ \frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt} = k \cdot v \]
Here, \( \frac{dx}{dt} = v \) is the velocity of the water jet.

### Step 6: Substitute back into the kinetic energy equation
Now substituting \( \frac{dm}{dt} \) back into the kinetic energy rate equation:
\[ \frac{d(KE)}{dt} = \frac{1}{2} v^2 \cdot (k \cdot v) \]
This simplifies to:
\[ \frac{d(KE)}{dt} = \frac{1}{2} k v^3 \]

### Final Result
Thus, the rate at which kinetic energy is being imparted to the water is:
\[ \frac{d(KE)}{dt} = \frac{1}{2} k v^3 \]

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