How Do You Calculate Kinetic Energy and Acceleration on a Loop-the-Loop Track?

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SUMMARY

The discussion focuses on calculating kinetic energy and acceleration for a block sliding on a loop-the-loop track. The block, with a mass of 1.3 kg, starts from a height of 52.0 m and moves along a track with a radius of 18.0 m. Key equations used include the conservation of mechanical energy, expressed as KE + PE = constant, where KE is kinetic energy (KE = 1/2mv^2) and PE is potential energy (PE = mgh). The participants successfully derive the kinetic energy at point A as 114.63 J and discuss the conditions for the block to complete the loop without leaving the track.

PREREQUISITES
  • Understanding of kinetic energy (KE = 1/2mv^2)
  • Knowledge of potential energy (PE = mgh)
  • Familiarity with the concept of conservation of mechanical energy
  • Basic principles of circular motion and centripetal acceleration
NEXT STEPS
  • Study the principles of conservation of energy in mechanical systems
  • Learn about centripetal acceleration and its calculations in circular motion
  • Explore the relationship between height and velocity in potential and kinetic energy contexts
  • Investigate real-world applications of kinetic energy calculations in engineering
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of energy conservation and motion dynamics.

  • #31
Can you show me how to set up the equations that I need?
 
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  • #32
bulldog23 said:
Can you show me how to set up the equations that I need?

You should be able to do this if you made it this far. You already set them up in parts a and b. What is the mechanical energy at P? What is the mechanical energy at A? Then, set these two quantities equal to each other, since you know they are equal, and you should be able to solve for h. As I said, you already did this in part a and b, but there you were just solving for something else. In simplest form then, the equation you want is:

ME_P=ME_A

Again, you should be able to find ME_P and ME_A, since you were able to do this in a) and b).
 
Last edited:
  • #33
So I do 1/2mv_1^2+mgy_1=1/2mv_2^2+mgy_2? And then I am solving for y_1 right?
 
Last edited:
  • #34
or do I just set mgh=204 and solve for h?
 
  • #35
bulldog23 said:
So I do 1/2mv_1^2+mgy_1=1/2mv_2^2+mgy_2? And then I am solving for y_1 right?

This is correct.
 

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