Question

An engine pumps water continuously through a hose. If the speed with which the water passes through the hose nozzle is \( v \), and if \( k \) is the mass per unit length of the water jet as it leaves the nozzle, what is the rate at which kinetic energy is being imparted to the water?

• A. \( 2kv^2 \)
• B. \( 2kv^3 \)
• C. \( 2kv^4 \)
• D. None

Hint:

To find the rate of energy imparted to a moving fluid, use the formula for kinetic energy and multiply by the velocity of the fluid.

Answer 1

The Correct Option is A

To determine the rate at which kinetic energy is being imparted to the water, we start by considering the kinetic energy of a small element of water as it leaves the nozzle. The mass of water ejected per unit time is given by:

\( \dot{m} = khv \)

where \( k \) is the mass per unit length and \( v \) is the velocity of water. The kinetic energy \( KE \) of a small element of mass per unit time moving with velocity \( v \) is given by:

\( KE = \frac{1}{2} \dot{m} v^2 \)

Substituting the expression for \( \dot{m} \):

\( KE = \frac{1}{2} (kv)v^2 = \frac{1}{2} kv^3 \)

Since the water is being continuously pumped, the rate at which kinetic energy is imparted is simply twice this expression:

\( \text{Rate of KE} = kv^3 \)

Thus, the power required (or the rate of kinetic energy imparted) by the engine to pump the water is given by the correct option:

\( 2kv^2 \)