Question

An engine pumps water continuously through a hose water leares the hose with a velocity \(\mathrm{V}\) and \(\mathrm{m}\) is the mass per unit length of the water Jet what is the rate at which kinetic energy is imparted to water. (A) \((1 / 2) \mathrm{mV}^{3}\) (B) \(\mathrm{mV}^{3}\) (C) \((1 / 2) \mathrm{mV}^{2}\) (D) \((1 / 2) \mathrm{mV}^{\alpha} \mathrm{V}^{2}\)

Short answer

The short answer is: The rate at which kinetic energy is imparted to water is \((1/2) mV^{3}\), where m is the mass per unit length of the water jet and V is the velocity.

Step-by-step solution

1. Recalling kinetic energy formula

The formula for kinetic energy (K) is given by: \[K = \frac{1}{2}mv^2\] Where m is the mass and v is the velocity. Since the question mentioned mass per unit length, we can modify the equation as follows: Let m' be the mass per unit length and L be the length of water segment inside the hose. Then, the total mass m of the water segment is: \[m = mL\]

2. Calculate the kinetic energy of the water segment

Substitute the expression for mass m in the kinetic energy formula: \[K = \frac{1}{2}(m'L)(V^2)\]

3. Calculate the rate of kinetic energy transfer

We are interested in finding the rate of change of kinetic energy with respect to time. To do this, we will differentiate the kinetic energy K with respect to time t. The first step is to differentiate the length L with respect to time t. From the problem's information, we know that the water jet moves at velocity V. As a result, the length of the water segment L changes with time (t) at the same rate. Therefore, the rate of change of length L with respect to time t is equal to the velocity V: \[\frac{dL}{dt} = V\] Now, we differentiate K with respect to time: \[\frac{dK}{dt} = \frac{1}{2}\left(\frac{d(m'LV^2)}{dt}\right)\] Here, m' is a constant, so we just need to differentiate the product (LV^2) with respect to time using the product rule: \[\frac{dK}{dt} = \frac{1}{2}\left(m' \left(\frac{dL}{dt}V^2 + L\frac{d(V^2)}{dt}\right)\right)\] Now, substitute the value of \(\frac{dL}{dt}\) into the equation: \[\frac{dK}{dt} = \frac{1}{2}\left(m' \left(VV^2 + L\frac{d(V^2)}{dt}\right)\right)\] Since the velocity V is constant, the derivative \(\frac{d(V^2)}{dt}\) is equal to zero. Therefore, the equation simplifies to: \[\frac{dK}{dt} = \frac{1}{2}m'V^3\] Comparing this result with the given options, we can see that the correct answer is: (A) \((1 / 2) \mathrm{mV}^{3}\)

Key concepts

Rate of Kinetic Energy Transfer

In physics, understanding the change of kinetic energy over time is crucial when analyzing systems in motion. In this exercise, we analyze the water jet being pumped through a hose. When the water exits the hose with a certain velocity, it possesses kinetic energy due to its motion. The rate at which this kinetic energy changes - or the **rate of kinetic energy transfer** - is what we're solving here.
To find this, we must differentiate the kinetic energy expression. Starting from the basic formula for kinetic energy, \[K = \frac{1}{2}mv^2\]we incorporate the mass per unit length and the constant velocity of the water jet. We consider how the length of water inside the hose changes over time as it transfers kinetic energy continuously. The rate of kinetic energy transfer becomes essential in understanding how efficiently the engine pumps energy into the water.
This leads us to a final rate result, \[\frac{dK}{dt} = \frac{1}{2}m'V^3\]indicating the continuous energy transfer for the moving water at velocity, providing understanding into real-world applications like fluid energy dynamics.

Mass per Unit Length

In physics problems, the term **mass per unit length** often comes up in scenarios involving ropes, strings, or, like our exercise, a water jet. It is a way to assign a mass density to a one-dimensional line of material, assisting in calculations where material flows or spans over a distance.
In this exercise, the water jet's mass per unit length, denoted as \(m'\), is a critical factor in determining the total mass involved in kinetic energy calculations. Assuming a certain length of the water in the hose has flowed in any given time, we determine its kinetic energy by multiplying this length by the mass per unit length to get total mass, \(m = m'L\).
This value then interacts with velocity as shown in kinetic energy formulas to understand how the mass affects energy transfer.

Differentiation of Kinetic Energy

The concept of **differentiation** is essential in physics, particularly when it comes to studying dynamic changes, like how kinetic energy evolves over time. Differentiating kinetic energy with respect to time helps us understand these changes.
Given the standard formula for kinetic energy and integrating it over the time dimension involves considering how both mass (through mass per unit length) and velocity contribute to energy change. Since velocity remains constant, the primary differentiating factor is the change in length of the water segment over time, which translates to a change in mass.
This differential understanding shows that constant velocity simplifies the differentiation process since the term \( \frac{d(V^2)}{dt} \) equals zero, leading to an expression like \[\frac{dK}{dt} = \frac{1}{2}m'V^3\]This outcome sheds light on how energy is influenced by both the rate of length change and kinetic factors, and ultimately supports engineering applications like fluid dynamics in pumps.