Looking for classes? Ksquare Career Institute, Bengaluru →
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?
A
$\frac{1}{2}m^2v^2$
B
$\frac{1}{2}mv^3$
C
$mv^3$
D
$\frac{1}{2}mv^2$
Detailed Solution
In time dt, a length $v\,dt$ of the water jet leaves the hose.
Mass of water leaving in time dt: $dM = m\times v\,dt$, so $\frac{dM}{dt} = mv$
Kinetic energy carried by this mass: $dK = \frac{1}{2}(dM)v^2$
Rate at which kinetic energy is imparted: $\frac{dK}{dt} = \frac{1}{2}\frac{dM}{dt}v^2 = \frac{1}{2}(mv)v^2$
$\frac{dK}{dt} = \frac{1}{2}mv^3$
Mass of water leaving in time dt: $dM = m\times v\,dt$, so $\frac{dM}{dt} = mv$
Kinetic energy carried by this mass: $dK = \frac{1}{2}(dM)v^2$
Rate at which kinetic energy is imparted: $\frac{dK}{dt} = \frac{1}{2}\frac{dM}{dt}v^2 = \frac{1}{2}(mv)v^2$
$\frac{dK}{dt} = \frac{1}{2}mv^3$