How Do You Calculate Velocity in a Horizontal Slingshot Setup?

Click For Summary

Homework Help Overview

The problem involves calculating the velocity of a mass in a horizontal slingshot setup, which consists of two identical springs and a cup holding a stone. The springs are initially at equilibrium and are stretched when the cup is pulled to a certain distance before being released.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of mechanical energy and how to calculate potential energy stored in the springs. There are questions about incorporating the equilibrium length of the springs into the calculations and how to determine the stretched length of the springs.

Discussion Status

Participants are exploring various approaches to calculate the stretched length of the springs and the corresponding potential energy. Some have provided guidance on using the Pythagorean theorem to find the stretched length, while others are clarifying the concept of equilibrium length. There is ongoing confusion regarding the calculations and the correct application of energy conservation principles.

Contextual Notes

Participants are working with specific values for spring constants and lengths, and there is a focus on understanding how these values affect the energy calculations. The discussion reflects uncertainty about the correct interpretation of the setup and the relationships between the variables involved.

J.live
Messages
95
Reaction score
0

Homework Statement


A horizontal slingshot consists of two light, identical springs (with spring constants of 74 N/m) and a light cup that holds a 1-kg stone. Each spring has an equilibrium length l0 = 49 cm. When the springs are in equilibrium, they line up vertically. Suppose that the cup containing the mass is pulled to x = 0.6 m to the left of the vertical and then released.




The Attempt at a Solution



ME= PE +KE

Initially PE = 1/2kx^2 KE= 0 --> ME = 1/2kx^2 +0 ---> 1/2(74)(.6)^2 +0 ?

Idk how to find the velocity. I am guessing we'll just use ME= 0+1/2mv^2 ?

Or, do I have to incorporate the equilibrium length into the equation somehow?

Any help will be appreciated.
 

Attachments

  • Screen shot 2010-10-04 at 3.19.41 PM.png
    Screen shot 2010-10-04 at 3.19.41 PM.png
    5.3 KB · Views: 501
Last edited:
Physics news on Phys.org
anyone ?
 
J.live said:
Or, do I have to incorporate the equilibrium length into the equation somehow?
Definitely. Figure out how much the springs are stretched from their equilibrium lengths when the cup is moved. (It's not simply 0.6 m.) Then your energy approach will work.
 
How do I calculate that ?:/

Edit: Pythagorean Theorem ?
 
J.live said:
How do I calculate that ?:/
How long are the stretched springs? How does that compare with their unstretched length?

Yes, Pythagorean theorem!
 
This is what I did

C^2 = .6^2 + .49 ^2 = .77 m

Total ME = 1/2(74)(.77)^2 +0 = 21. 97 J ?

That seems off. Am I suppose to subtract and add the x and y components before applying the Pythagorean Theorem ?

If so then isn't y = 0 since they are in opposite directions? x = .8 +.8 = 1.6 ?

I am still kind of confused.

Edit- I just tried it that makes no sense whatsoever. :/ What am i doing wrong ?
 
Last edited:
When I subtract the values I am getting the right answer. C^2= .64^2 - .48^2 ?
 
J.live said:
This is what I did

C^2 = .6^2 + .49 ^2 = .77 m
You correctly found the stretched length of the springs. So how much did they stretch compared to their unstretched length? That elongation will tell you the amount of stored elastic potential energy.
 
I don't understand which value you are referring to exactly as unstretched length?

.49 m?
 
Last edited:
  • #10
J.live said:
I don't understand which value you are referring to exactly as unstretched length?

.49 m?
Yes, what they are calling the equilibrium length or l0. To find the energy stored in the stretched springs, you need to know how far they are stretched from their equilibrium positions.
 
  • #11
So, will it be .77 -.49 = .28 ?
 
  • #12
J.live said:
So, will it be .77 -.49 = .28 ?
Good.
 
  • #13
Now I plug it in the equation

ME = 1/2kx^2 +0 --> 1/2 74 (.28)^2= 2.9 J ?
 
  • #14
J.live said:
Now I plug it in the equation

ME = 1/2kx^2 +0 --> 1/2 74 (.28)^2= 2.9 J ?
That's the energy stored in each spring.
 
  • #15
ME= 2*(1/2 ) 74 (.28)^2 = 5.99 j
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
1K
Replies
7
Views
1K
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K