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Question: An engine pumps water continuously through a hole. The speed with which water passes through a hole ...

An engine pumps water continuously through a hole. The speed with which water passes through a hole nozzle is vv, and kk 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.

Explanation

Solution

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

Formula used:
KE=12mv2KE=\dfrac{1}{2}mv^{2} and rate of KE=ddt(KE)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=12mv2KE=\dfrac{1}{2}mv^{2}, where mm is mass of the object and vv is the velocity of the object.

Given that, k=masslength=dmdxk=\dfrac{mass}{length}=\dfrac{dm}{dx} and vv is velocity.
Rate of KE=ddt(KE)=12(dmdt)v2=12(dmdxdxdt)v2=12kv3KE=\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 12kv3\dfrac{1}{2}kv^3
Alternatively, if it takes tt time for water to flow through length ll with the velocity vv,
Then we know t=lvt=\dfrac{l}{v}
Mass mm of water that flows in time t=klt=kl
Then KE\text{KE} per unit time is given by =12mv2t=\dfrac{1}{2}\dfrac{mv^{2}}{t}
=12klv2lv=12kv3=\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 12kv3\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. dmdt=(dmdxdxdt)\dfrac{dm}{dt}=(\dfrac{dm}{dx}\cdot\dfrac{dx}{dt})