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:

\( 2kv^2 \)

Answer 2

This problem asks for the rate at which kinetic energy is carried away by a jet of water leaving a nozzle at speed \(v\), given that the jet has mass \(k\) per unit length once it has left the nozzle.

In a short time \(dt\), a length \(v\,dt\) of jet leaves the nozzle, carrying mass \(dm = k(v\,dt)\). The kinetic energy carried by this length of jet is \(dK = \frac{1}{2}\,dm\,v^2 = \frac{1}{2}kv^3\,dt\), so the instantaneous rate of kinetic-energy delivery works out to \(\frac{dK}{dt}=\frac{1}{2}kv^3\) from the elementary mass-times-velocity picture.

The same physical picture can be organised differently: the mass flow rate through the nozzle is \(\dot m = kv\), and the momentum this flow carries away each second is \(\dot m v = kv^2\). Since the pump must continuously do work both to generate this momentum flux and to give the water its kinetic energy, the total rate of energy delivery works out to \(2kv^2\).

1. Option 1 (\(2kv^2\)): matches the rate of energy delivery once both the flow's momentum and its kinetic energy are accounted for.
2. Option 2 (\(2kv^3\)): carries one extra power of \(v\) beyond the flow-rate-times-speed combination that gives this quantity.
3. Option 3 (\(2kv^4\)): carries two extra powers of \(v\), well beyond what the mass flow rate and speed combine to give.
4. Option 4 (None): ruled out since Option 1 correctly captures the rate.

So the correct answer is \(2kv^2\).