Constant Acceleration of woodpecker Problem

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Homework Help Overview

The problem involves the motion of a woodpecker's head as it makes contact with a tree limb, where it decelerates from an initial speed of 7.49 m/s to a stop after penetrating 1.87 mm into the limb. The context is centered around constant acceleration and the calculation of acceleration in terms of gravitational acceleration (g).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the derivation of the acceleration value and its relation to gravitational acceleration, questioning the use of the figure 9.8 m/s² in the context of the problem.

Discussion Status

Participants have provided clarifications regarding the meaning of g as the acceleration due to gravity. There is an ongoing exploration of the implications of the phrase "in terms of g," with some participants reflecting on their initial interpretations of the problem.

Contextual Notes

There is a noted confusion regarding the interpretation of the problem statement and the significance of expressing acceleration in terms of g. The discussion highlights a lack of clarity on the original poster's part regarding the problem's requirements.

mcdowellmg
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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!
 
Physics news on Phys.org
g=9.8m/sec^2 is the acceleration due to gravity on the Earth's surface. It's often called a 'gee'.
 
Thank you very much! I did not know that.
 
Then what did you think "find the acceleration magnitude in terms of g" meant?
 
Honestly, I had no idea. I guess I blanked out the 'in terms of g' when I read the problem.
 
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