Calculate kinetic energy transferred

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The discussion revolves around calculating the kinetic energy transferred to a fluid during an experiment, with specific focus on questions b i, b ii, and b iii. Participants emphasize the importance of using precise values in calculations to avoid rounding errors and suggest using the kinetic energy formula EK = 1/2 m v^2 with the correct mass of the ejected fluid. The conversation also touches on the need to understand the projectile motion of the fluid, clarifying that the initial and final velocities must be treated as vectors with appropriate signs. Ultimately, the key takeaway is that the mass of the fluid ejected can be determined by the difference in mass before and after the experiment, leading to the calculation of kinetic energy. Proper application of equations of motion is essential for accurate results.
  • #31
gneill said:
Well, I think you meant parabola rather than hyperbola, but yes, that's about it.
Oh yes that's what I meant! Thank you so much for helping me this evening, we got there in the end :) I really appreciate it!
 
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  • #32
gneill said:
Well, I think you meant parabola rather than hyperbola, but yes, that's about it.
Wait, but the vertical velocity does change because of acceleration due to gravity?
 
  • #33
Nicaragua said:
Wait, but the vertical velocity does change because of acceleration due to gravity?
Yes.
 
  • #34
gneill said:
Yes.
I am really sorry but I am re-visiting this question and do not understand it. Can you please explain to me how you work out the kinetic energy transferred when leaving the bottle? Do you use 1/2 m v^2 with v = the final velocity of the fluid (7.5 m/s^2)?
 
  • #35
Nicaragua said:
I am really sorry but I am re-visiting this question and do not understand it. Can you please explain to me how you work out the kinetic energy transferred when leaving the bottle? Do you use 1/2 m v^2 with v = the final velocity of the fluid (7.5 m/s^2)?
Not the final velocity, the initial velocity that the fluid has as it leaves the bottle. That's the 7.5 m/s value. Yes, you apply that speed and the total mass that's been given that speed to the kinetic energy formula.

The final velocity would be the speed of the fluid as it hits the ground at the end of its trajectory.
 
  • #36
gneill said:
Not the final velocity, the initial velocity that the fluid has as it leaves the bottle. That's the 7.5 m/s value. Yes, you apply that speed and the total mass that's been given that speed to the kinetic energy formula.

The final velocity would be the speed of the fluid as it hits the ground at the end of its trajectory.
So is the total mass that has been given the velocity of 7.5 ms^-2 2.24 kg?
 
  • #37
Nicaragua said:
So is the total mass that has been given the velocity of 7.5 ms^-2 2.24 kg?
Review posts #12 and #13.
 

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