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Constant Acceleration of woodpecker Problem

  • Thread starter mcdowellmg
  • Start date
  • #1
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Homework Statement


The head of a woodpecker is moving forward at a speed of 7.49 m/s when the beak makes first contact with a tree limb. The beak stops after penetrating the limb by 1.87 mm. Assuming the acceleration to be constant, find the acceleration magnitude in terms of g.

Know: velocity (v) = 0, initial velocity (v0) = 7.49m/s, and displacement during acceleration (x-x0) is 1.87x10^-3m.


Homework Equations



v^2 = v0^2 + 2a(x-x0)


The Attempt at a Solution



I was given the solution, but do not understand where a certain figure comes from. They plug the known quantities into the equation to get 0^2 = 7.49^2 + 2a(1.87*10^-3m), which algebraically reduce to a = -1.500 x 10^4m/s^2.
However, the next step loses me. They divide by g = 9.8m/s^2, take the absolute value, and that gives the magnitude of the head's acceleration is a = (1.53x10^3).

Where does that 9.8m/s^2 come from?! Thanks!
 

Answers and Replies

  • #2
Dick
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g=9.8m/sec^2 is the acceleration due to gravity on the earths surface. It's often called a 'gee'.
 
  • #3
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Thank you very much! I did not know that.
 
  • #4
HallsofIvy
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Then what did you think "find the acceleration magnitude in terms of g" meant?
 
  • #5
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Honestly, I had no idea. I guess I blanked out the 'in terms of g' when I read the problem.
 
  • #6
tiny-tim
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