태권도 물리: Lesson 2 – velocity and force

First, we start with v=\sqrt[]{\frac{2K}{m}}

  • Let m = mass of the fist and arm of a 45 kg (100 pound) martial artist, approximately 10 kg (22 pounds).
  • Let K = kinetic energy solved in lesson 1, E_{arm}=\frac{1}{2}(10)(5)^2-(5)(9.8)=125J-49J=76J

Next, plug and chug: v=\sqrt[]{\frac{2(76)}{10}}=3.9m/s

This shows that the initial assumption of 5 m/s was close enough. We can re-visit lesson 1 and show the difference in energy and force.

Insert corrected values: E_{arm}=\frac{1}{2}(10)(3.9)^2-(5)(9.8)=76J-49J=27J

Note that with a 1.1 m/s change in velocity, there is about a 50J change in energy delivered, but the force applied does not change! Force is a product of mass and acceleration, not velocity! To state that again, as long as I set my initial conditions such that the initial guess on velocity is close, I can back-derive the actual velocity by focusing on the key variables of mass and acceleration. The work shown above has been abbreviated. I leave the full derivation up to the martial arts physicist who happens upon this lonely blog.

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6 thoughts on “태권도 물리: Lesson 2 – velocity and force

  1. Pingback: Week 36 of 213: I want to know God’s thoughts. The rest are only details. « Aerospace Cubicle Engineer (ACE)

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  3. Pingback: 태권도 물리: Lesson 1 – energy and force | Aerospace Cubicle Engineer (ACE)

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