Question

An engine pumps out water continuously through a hose with a velocity $v$. If $m$ is the mass per unit length of the water jet, the rate at which the kinetic energy is imparted to water is

• A. $\frac{1}{2}m{v}^2$
• B. $\frac{1}{2}m{v}^3$
• C. $\frac{1}{2}{m}^2{v}^2$
• D. $m{v}^3$
• E. $\frac{1}{4}m{v}^3$

Answer 1

Answer: (B)

$\frac{1}{2}m{v}^3$

Rate of kinetic energy
$KE=\frac{d}{dt}\frac{1}{2}M{v}^2$
But $M=mL$
$KE=\frac{1}{2}\frac{dm}{dt}L{v}^2=\frac{1}{2}m{v}^3$