Calculating k in Hooke's Law: Understanding Archery and Projectile Motion

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    Hooke's law Law
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Homework Help Overview

The discussion revolves around calculating the spring constant (k) in the context of Hooke's Law, specifically applied to an archery scenario involving projectile motion. The original poster describes a situation where an archer pulls back a bowstring and seeks to determine k based on the arrow's weight and distance traveled.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between potential energy stored in the bow and the kinetic energy of the arrow. Questions arise regarding the initial velocity required for the arrow to reach a specified distance and the optimal angle for launch. Some participants clarify equations and discuss the implications of energy conservation in the context of the problem.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on how to approach the problem. There is a focus on determining the necessary initial velocity for the arrow and understanding the energy transformations involved. Multiple interpretations of the problem are being explored, particularly regarding the energy equations and their application.

Contextual Notes

Participants note the importance of neglecting air resistance and the assumption that the arrow's initial velocity is critical for achieving the desired range. There is also mention of the potential energy associated with the bowstring's displacement.

aatgomez
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In preparing to shoot an arrow, an archer pulls a bowstring back 20 cm=.2 m. The arrow can reach a horizontal distance of 300 m and the arrow weighs 200 g. Find k.

m=.2 kg
W=1/2kd final- 1/2kd initial
Fs=-kx

Not really sure where to begin, and I've been trying for a while now. :confused:
 
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There are squares missing in your second equation.

To shoot 300m, how fast does the arrow have to be initially? And what is the ideal angle to shoot the arrow, neglecting air resistance?
 
oh yes, sorry W=1/2kd^2 final - 1/2kd^2 initial
It would have to be 0 initially and I'm thinking 45 degrees?
 
45 degrees is right. With "initially", I meant the flight path, so directly after it leaves the bow. If it is too slow, it won't make 300 meters, obviously.
 
Ok, so if I find the initial velocity and the final velocity as if I were doing projectile problem then I can find W=1/2mvf^2 - 1/2mvi^2 then I could set W equal to 1/2kd^2 final - 1/2kd^2 initial and find k. Is that correct?
 
No. The energy from the spring does not get "used up" in the flight. It is still in the arrow when it hits the ground.

The velocity is the key connection here.
 
I see that there's potential energy before the arrow is shot. That's where the .2 m come in. But I still don't know how to solve it. A little more help please?
 
Find the velocity the arrow needs to fly 300m. Once you have that, the rest should follow from that velocity. Note that is the same hint I gave in post #2.
 
Right I realize that, that's what I don't know how to find.
 
  • #10
The arrow is in free fall. Maybe the opposite problem is easier: If you know the initial velocity and angle, how far will it go?
I'm quite sure you have solved a problem of this type before in your course or as homework.
 

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