Calculating Changes in Water Speed with Kinetic Energy Equations

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SUMMARY

The discussion centers on calculating the change in water speed when the kinetic energy increases by a factor of 2.25. The key equation used is the kinetic energy formula, Ek = 0.5mv². By setting Ek(new) to 2.25 times Ek(old), the relationship between the new and old speeds is derived. The final calculation shows that the new speed is the square root of 4.5, indicating a speed change factor of approximately 2.12.

PREREQUISITES
  • Understanding of kinetic energy equations, specifically Ek = 0.5mv²
  • Basic algebra skills for manipulating equations
  • Familiarity with square roots and their application in physics
  • Knowledge of units of measurement for mass and velocity
NEXT STEPS
  • Study the derivation of kinetic energy equations in physics
  • Learn about the implications of energy conservation in fluid dynamics
  • Explore practical applications of kinetic energy calculations in engineering
  • Investigate the relationship between kinetic energy and other forms of energy
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Students in physics, educators teaching mechanics, and anyone interested in the principles of energy transformation in fluid dynamics.

McKeavey
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Homework Statement


It was on a test, not sure if I remember it to the exact word but it went something like..
If the Kinetic Energy of flowing water increases by a factor of 2.25, what is the factor that the speed changes by.
Something like that :S


Homework Equations





The Attempt at a Solution


I said something like..

The square root of 4.5/m
 
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Try it like this:

Ek(new)/Ek(old) = 2.25

Ek(new)=0.5mvnew2

Ek(old) = ?

Now just sub it into the equation.
 
Say KE=1 originally. It would become 2.25 if increased by a factor of 2.25. With the Equation KE=.5mv^2, you can isolate the v^2 to get v^2= KE/(.5m). Just replace the values and you would get that answer. 2.25/.5= 4.5 which becomes the square root of 4.5 after you square root velocity
 

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