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) \((1 / 2) \mathrm{mV}^{\alpha} \mathrm{V}^{2}\)

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: \[m = mL\]

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

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) \((1 / 2) \mathrm{mV}^{3}\)