An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

\(\frac{1}{2}mv^3\)

mass per unit length : m

Length covered by water in 1 second is v

Amount of mass passing per second = mv

Rate of K.E. = $\frac{d(mv^2)}{2dt}= \frac{dm}{2dt}v^2= \frac{mv}{2}v^2 = \frac{mv^3}{2}$