With what speed does the stone leave the slingshot?

  • Thread starter Thread starter imdumb
  • Start date Start date
  • Tags Tags
    Slingshot Speed
Click For Summary

Homework Help Overview

The discussion revolves around a problem involving a slingshot, specifically focusing on the potential energy stored in the rubber bands and the speed of a stone when released. The subject area includes concepts from mechanics and energy, particularly related to elastic potential energy and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between force, spring constant, and displacement in the context of Hooke's law. There are attempts to calculate potential energy and questions about the correct values to use in the equations.

Discussion Status

Some participants have provided guidance on using the correct formulas and understanding the parameters involved, while others are questioning the accuracy of previous calculations and clarifying the dimensions used in the problem. Multiple interpretations of the problem setup are being explored.

Contextual Notes

There is a noted confusion regarding the units of measurement, specifically the conversion from meters to centimeters, and the distinction between the spring constants of the individual rubber bands versus the total system. Additionally, the original poster expresses uncertainty about how to initiate the problem-solving process.

imdumb
Messages
3
Reaction score
0
A slingshot consists of a light leather cup containing a stone that is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of the bands 1.0 cm. A)What is the potential energy in the two bands together when a 47g stone is placed in the cup and pulled back.12 m from the equilibrium position? B) With what speed does the stone leave the slingshot? I'm not sure where to start!
 
Physics news on Phys.org
you have to consider the rubber band has happy Hookian springs. I.e. such that F = kx.

By "It takes a force of 15 N to stretch either one of the bands 1.0 cm", what they mean is that if you exert a force of 15N on one of the rubber band, you're not going to be able to stretch it farther than 1cm. At this point, equilibrium will have been attained. this means that at x = 1cm, the force of the rubber band is of 15N too. From this, you get k = 15 [N/cm].

You go from there.
 
Last edited:
so... then would 7.5 be the correctanswer to part A?
 
Nope. How did you get that answer?

(Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)
 
Last edited:
I used the equation PE=.5kx^2and put 15 infor k and 1 in for x.
 
That's the right idea, but pay more attention.

The actual problem talks about a rock that is pulled back.12 m (12 cm, not 1cm). Moreover, 15 [N/m] is the spring constant of ONE rubber band. You got two.
 

Similar threads

Replies
3
Views
16K
  • · Replies 4 ·
Replies
4
Views
14K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 2 ·
Replies
2
Views
23K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
14
Views
3K
Replies
15
Views
6K
  • · Replies 16 ·
Replies
16
Views
4K