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

  • Thread starter Thread starter Hughey85
  • Start date Start date
AI Thread Summary
To calculate the speed of an arrow released from a bow modeled as a spring, use the spring potential energy formula W = 0.5kx^2, where k is the spring constant (80 N/m) and x is the draw length (0.51 m). The mechanical energy conservation principle states that the initial potential energy in the bow converts entirely into kinetic energy at release. The weight of the arrow can be disregarded for this calculation, as the effect of gravity is minimal over the short distance of 51 cm. By equating the potential energy to kinetic energy, you can solve for the arrow's speed upon release.
Hughey85
Messages
14
Reaction score
0
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:
 
Physics news on Phys.org
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.
 
Back
Top