Two springs and the energy question

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coldpay
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Homework Statement



A horizontal slingshot consists of two light, identical springs (with spring constants of 24.1 N/m) and a light cup that holds a 1.21-kg stone. Each spring has an equilibrium length of 50 cm. When the springs are in equilibrium, they line up vertically. Suppose that the cup containing the mass is pulled to x = 0.7 m to the left of the vertical and then released. Determine

a) the system’s total mechanical energy.


b) the speed of the stone at x = 0.




Homework Equations





The Attempt at a Solution

 
Attachments
  • Diagram of a horizontal slingshot: two springs angled from a cup on the left to fixed walls at top and bottom, with a stone in the cup, force F pointing left, and dimensions 0.5 m above and below the centerline and x = 0.7 m to the cup.
    P047figure.png
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welcome to pf!

hi coldpay! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
i know the the total mechanical energy is mgh plus 0.5kx^2 but i can't find the answer.

answers
a)3.127 J
b)2.274 m/s
 
coldpay said:

Homework Statement



A horizontal slingshot consists of two light, identical springs (with spring constants of 24.1 N/m) and a light cup that holds a 1.21-kg stone. Each spring has an equilibrium length of 50 cm. When the springs are in equilibrium, they line up vertically. Suppose that the cup containing the mass is pulled to x = 0.7 m to the left of the vertical and then released. Determine

a) the system’s total mechanical energy.


b) the speed of the stone at x = 0.




Homework Equations





The Attempt at a Solution


coldpay said:
i know the the total mechanical energy is mgh plus 0.5kx^2 but i can't find the answer.

answers
a)3.127 J
b)2.274 m/s

I don't think the mgh term will change by pulling back the slingshot. Calculate how much energy is stored in the springs as they are pulled from their equilibrium postition back to the position shown. Show us your work please.
 
X=√0.5^2+0.7^2=0.86

0.5kx^2=0.5*48.2*(0.86)^2=17.34 i found this but i am not sure if the spring is parallel or serial(i take it parallel)

This is all i can do.I don't know the rest of the question.
 
coldpay said:
X=√0.5^2+0.7^2=0.86

0.5kx^2=0.5*48.2*(0.86)^2=17.34 i found this but i am not sure if the spring is parallel or serial(i take it parallel)

This is all i can do.I don't know the rest of the question.

In the spring equation, the "X" is meant to be the amount of stretch, not the overall length.