Calculate Arrow Speed w/ Spring Constant & Draw Length - Hughey

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Homework Help Overview

The discussion revolves around calculating the speed of an arrow released from a bow modeled as a spring, with a given spring constant and draw length. The problem involves concepts from mechanics, specifically the conservation of mechanical energy and the relationship between potential and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the spring potential energy equation and the conservation of mechanical energy. Questions arise about the relevance of gravitational potential energy and whether the weight of the arrow affects the calculation.

Discussion Status

There is ongoing exploration of the problem with various approaches being suggested, including focusing on spring potential energy and the implications of mechanical energy conservation. Some participants question the significance of gravitational effects, while others suggest simplifying assumptions such as horizontal firing.

Contextual Notes

Participants note the absence of information regarding the shooting angle and the potential neglect of gravitational effects due to the short distance involved in the draw length.

Hughey85
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Help pls! Here is the question:

"Assuming you have a bow that behaves like a spring with a spring constant of 80 N/m and you pull it to a draw of 51 cm, to the nearest tenth of a m/s, what is the speed of the 81-gram arrow when it is released?"

What equations do you use? I'be been trying W=.5kx^2, but don't know where to go from there. Pls. help me think through this! Thanks!

Hughey :smile:
 
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Mechanical energy is conserved. Does that help you?
 
No, not really. If you found the Force of the "draw-back" bow, could you figure out the speed of the arrow? Because it says when the arrow is released, does that mean you don't need to take into account the weight of the arrow?
 
In the Conservartion of Mechanical Energy

\Delta K + \Delta \Omega = 0

The potentials energy not necessarily has to be both, it could be one, like Spring-Potential Energy \frac{1}{2} kx^2, and/or Gravitational Potential Energy mgh
 
You are quite correct in assuming that in the general, non-horizontal firing of the arrow, a change in potential energy due to gravity will be present.
However, do you think a change of potential gravitational energy of a distance over maximally 51 cm is significant compared to the potential energy stored in the drawn-back string?
Disregard the the effect of gravity, it shouldn't matter.
(That's one of the reasons why they haven't provided you with information concerning the shooting angle..)
 
Alternatively, solve the problem under the EXPRESS condition that you are aiming horizontally.
 
I'm still a little confused. What value do you have to find first? If you use Spring-Potential Energy equation, how would you find the speed of the arrow?
 
Conservation of Mechanical Energy

\Delta K + \Delta \Omega = 0

At any points of the system, the mechanical energy will be the same. You can find speed with Kinetic energy.
 
Last edited:
Initially, all you've got is potential energy; finally, all you've got is kinetic energy.
 

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