Finding acceleration and time in force problems

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Homework Help Overview

The discussion revolves around a physics problem involving the acceleration and time of a woodpecker's head during impact with a tree. The problem includes given values such as speed, mass, and force, and requires participants to find acceleration, depth of penetration, and time to stop.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of Newton's second law (F=ma) to find acceleration and discuss unit conversions between km/s² and m/s².
  • Some participants question the appropriateness of the units used for acceleration and depth, while others suggest checking signs for deceleration.
  • There is a discussion about the relationship between distance, acceleration, and time, with references to relevant equations.

Discussion Status

Participants have made attempts to calculate acceleration and have identified issues with unit consistency. Some have successfully calculated the depth of penetration but are still working on the time calculation. Guidance has been provided regarding the need to match units and reconsider the equations used.

Contextual Notes

There is an ongoing discussion about the correct interpretation of the problem's parameters, including the need for consistent units and the implications of initial velocity in the equations being used.

a18c18
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Homework Statement



The speed of the head of a redheaded woodpecker reaches 3.7 m/s before impact with the tree. The mass of the head is 0.060 kg and the average force on the head during impact is 6.1 N.

(a) Find the acceleration of the head (assuming constant acceleration).
km/s2

(b) Find the depth of penetration into the tree.
cm

(c) Find the time it takes for the head to come to a stop.


Homework Equations



F=ma
F=mv^2/2d

The Attempt at a Solution



For a I tried to used the formula F=ma and divided the force by the mass to find the acceleration or 6.1/0.060 but this did not work.

I got the answer for b by using F=mv^2/2d and got 6.7cm which was correct.

In order to get the answer for c I think I need the answer for a.
 
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Are you sure for answer a that you are using the right units for the answer. km/s^2?
 
I tried 6.1/.060 and got 101.6667 m/s^2. Then I divided 101.6667 by 1000 since the answer needs to be km/s^2 and got .1016 but neither of these answer were correct.
 
a18c18 said:

Homework Statement



The speed of the head of a redheaded woodpecker reaches 3.7 m/s before impact with the tree. The mass of the head is 0.060 kg and the average force on the head during impact is 6.1 N.

(a) Find the acceleration of the head (assuming constant acceleration).
km/s2

(b) Find the depth of penetration into the tree.
cm

(c) Find the time it takes for the head to come to a stop.

Homework Equations



F=ma
F=mv^2/2d

The Attempt at a Solution



For a I tried to used the formula F=ma and divided the force by the mass to find the acceleration or 6.1/0.060 but this did not work.

I got the answer for b by using F=mv^2/2d and got 6.7cm which was correct.

In order to get the answer for c I think I need the answer for a.

Is that really km/s2?

Also the acceleration is really deceleration to zero. Perhaps you have the wrong sign?

Part b) is found by the V2/(2*a) = x relationship.

Part c) is found from x = 1/2 a*t2
 
Okay I switched the signs for the acceleration and -.1016 was correct so thank you very much LowlyPion! but I still could not get c to work. Is x in c supposed to be the answer from b? I tried 6.7=.5*-.1016t^2 but got the wrong answer.
 
Make sure your units match. You are putting km/s^2 in for a, but the putting distance x in cm. Either change a or x to match the other's units.
 
a18c18 said:
Okay I switched the signs for the acceleration and -.1016 was correct so thank you very much LowlyPion! but I still could not get c to work. Is x in c supposed to be the answer from b? I tried 6.7=.5*-.1016t^2 but got the wrong answer.

Yes the answer from b is the distance. But whoa. wait. You have km/s2 acceleration and cm of distance. Please resolve that.
 
Also the equation needs to have initial Velocity added to it:
X = V0*t + .5*a*t^2
Am I right?
 
Lazyshot said:
Also the equation needs to have initial Velocity added to it:
X = V0*t + .5*a*t^2
Am I right?

If it's uniform deceleration forget the initial velocity. Treat it as accelerating from rest. It's the same time over that distance.
 

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