Calculate kinetic energy transferred

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SUMMARY

The forum discussion focuses on calculating the kinetic energy transferred to a fluid during an experiment involving projectile motion. The kinetic energy (KE) is calculated using the formula EK = 1/2 x m x v^2, where the mass of the fluid ejected is 1.45 kg and the velocity is 7.5 m/s, resulting in a KE of 40.781 Joules. Participants emphasize the importance of using unrounded values for intermediate calculations to avoid significant errors and discuss the relationship between initial and final velocities in vertical motion.

PREREQUISITES
  • Understanding of kinetic energy calculations using EK = 1/2 x m x v^2
  • Knowledge of projectile motion principles
  • Familiarity with conservation of energy concepts
  • Ability to apply equations of motion (EOM) for vertical and horizontal components
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  • Research the "Range Equation" for projectile motion calculations
  • Study the effects of rounding errors in physics calculations
  • Learn about vector components in projectile motion
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Students studying physics, particularly those focusing on mechanics and energy transfer, as well as educators seeking to enhance their teaching methods in projectile motion and energy conservation.

  • #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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