"Calculating the Velocity of an Arrow Hitting Hay

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Homework Help Overview

The problem involves calculating the initial velocity of an arrow that penetrates a bale of hay, given its mass, penetration distance, and the average frictional force acting on it. The subject area relates to dynamics and energy principles, particularly the kinetic energy-work theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the work done by the frictional force and its relation to the kinetic energy of the arrow. There are questions about the application of the kinetic energy-work theorem and whether the work done remains negative when the arrow comes to a stop.

Discussion Status

Some participants have provided guidance on using the work-energy principle, noting that the work done by the frictional force is negative due to the decrease in kinetic energy. There is an ongoing exploration of the implications of this theorem in the context of the problem.

Contextual Notes

Participants are navigating through the implications of the given values and the assumptions related to the frictional force and the motion of the arrow. There is a focus on understanding the relationship between work and energy without reaching a definitive conclusion.

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Homework Statement


A 1kg arrow hits a bail of hay and penetrates 47.0cm. The average frictional force of the hay on the arrow is 2500.0N. What was the original velocity of the arrow?


Homework Equations


Ffr = uk x Fn


The Attempt at a Solution


I'm completely lost, can someone please walk me through this?
 
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Well we know that,

W = fdcos(theta).

cos(theta) is going to go to 1 so,

W = fd.

We're given then average force and the distance so we can solve for the work done.

After you've done that I want you to think about the kinetic energy - work theorem ;)
 
is the kinetic energy-work theorem net work = change in kinetic energy?
so then, w = 1/2mvf^2-1/2mvi^2, but vf is 0 because the arrow is stopped in the hay so really w = -1/2mvi^2. does that stay negative?
 
lking226 said:
is the kinetic energy-work theorem net work = change in kinetic energy?
so then, w = 1/2mvf^2-1/2mvi^2, but vf is 0 because the arrow is stopped in the hay so really w = -1/2mvi^2. does that stay negative?

Yes, the work done is negative, because the kinetic energy decreaes (in this case all the way to zero, so that the change in kinetic energy is just the negative of how much there was to start with).
 
got it, thanks guys!
 

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