Question

An engine pumps water continuously through a hole. The speed with which water passes through a hole nozzle is $v$, and $k$ is the mass per unit length of the water jet as it leaves the nozzle. Find the rate at which kinetic energy is being imparted to the water.

Answer 1

Hint: Objects in motion possess kinetic energy. It is given that $k=\dfrac{mass}{length}=\dfrac{dm}{dx}$ and $v$ is velocity. To find the rate of change in kinetic energy, we must differentiate the \[\text{KE}\] with respect to time.

Formula used:
$KE=\dfrac{1}{2}mv^{2}$ and rate of $KE=\dfrac{d}{dt}(KE)$

Complete step-by-step answer:
We know from the law of inertia that work needs to be done to move an object which is at rest. This energy is transferred as kinetic energy of the object. We also know that, the kinetic energy is the energy possessed by the objects in motion. It is also defined as the work required by the object to accelerate. It is a scalar quantity, which is independent of the direction and is always positive.
It is given by $KE=\dfrac{1}{2}mv^{2}$, where $m$ is mass of the object and $v$ is the velocity of the object.

Given that, $k=\dfrac{mass}{length}=\dfrac{dm}{dx}$ and $v$ is velocity.
Rate of $KE=\dfrac{d}{dt}(KE)=\dfrac{1}{2}(\dfrac{dm}{dt})v^{2}=\dfrac{1}{2}(\dfrac{dm}{dx}\cdot\dfrac{dx}{dt})v^{2}=\dfrac{1}{2}kv^{3}$
Hence, the rate at which kinetic energy is being imparted is $\dfrac{1}{2}kv^3$
Alternatively, if it takes $t$ time for water to flow through length $l$ with the velocity $v$,
Then we know $t=\dfrac{l}{v}$
Mass $m$ of water that flows in time $t=kl$
Then \[\text{KE}\] per unit time is given by $=\dfrac{1}{2}\dfrac{mv^{2}}{t}$
$=\dfrac{1}{2}\dfrac{klv^{2}}{\dfrac{l}{v}}=\dfrac{1}{2}kv^{3}$
Hence, the rate at which kinetic energy is being imparted is $\dfrac{1}{2}kv^3$

Note: Kinetic energy is the work required by the object to accelerate. It is a scalar quantity, which is independent of the direction and is always positive. When differentiating KE with respect to time, we need to manipulate a little. $\dfrac{dm}{dt}=(\dfrac{dm}{dx}\cdot\dfrac{dx}{dt})$