The law of conservation of energy problem

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SUMMARY

The problem involves calculating the speed of water at the top of Della Falls, Canada, using the law of conservation of energy. Given a change in elevation of 440 m and a speed of 33 m/s after falling 12% of the height, the correct speed at the top is determined to be 5.0 m/s, as stated in the textbook. The equations used include kinetic energy (Ek = 1/2 mv^2) and potential energy (Ep = mgh), which lead to the conclusion that the initial calculations were incorrect due to misapplication of energy conservation principles.

PREREQUISITES
  • Understanding of kinetic energy and potential energy equations
  • Familiarity with the concept of conservation of energy
  • Basic knowledge of gravitational acceleration (9.81 m/s²)
  • Ability to perform algebraic manipulations and solve equations
NEXT STEPS
  • Review the principles of conservation of energy in physics
  • Practice problems involving gravitational potential energy and kinetic energy
  • Learn about energy transformations in fluid dynamics
  • Explore the effects of air resistance and fluid friction on energy calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of energy problems in real-world scenarios.

dmitrip
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hello, I been doing physics homework and I came across this problem that i think i know how to do but for some reason i cannot get the right answer! any help will be very appreciated
thanks a lot:)

Homework Statement



The highest waterfall in Canada is the Della Falls in B.C. with a change in elevation of 440 m m. When the water has fallen 12% of its way to the bottom, its speed is 33 m/s. Neglecting air resistance and fluid friction, determine the speed of the water at the top of the waterfall.

answer in the book: 5.0 m/s

Homework Equations



Ek = 1/2 mv^2

Ep = mgh


The Attempt at a Solution



This is what i tried,

i found what 12% of 440 m is, and it ended up to equal 52.8 so i subtracted it by 440 m to get 387.2 m as the height after the water has fallen 12% of its way to the bottom.

Ek1 + Ep1 = Ek2 + Ep2

(masses cancel out) and we are left with

1/2 v^2 + gh = 1/2 v^2 + gh

1/2 v^2 + (9.81)(440) = 1/2 (33)^2 + (9.81)(387.2)

and i get v= 66 m/s (dont think it is right but it is possible that the book might be wrong)
 
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Draw a sketch of the waterfall, and everything the question told you about the waterfall. Then mark on the value you got for the speed of the water at the top of the waterfall, and the value that the book got.

Once you've done this, you'll be able to figure out which answer is definitely wrong. Then you'll know whether you need some help with your calculation or not.
 

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