Posted by **Anonymous** on Friday, December 14, 2012 at 10:27am.

A horizontal slingshot consists of two light, identical springs (with spring constants of 37.7 N/m) and a light cup that holds a 1.37-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.

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**bobpursley**, Friday, December 14, 2012 at 11:14am
Is there a question here?

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**Elena**, Friday, December 14, 2012 at 12:47pm
(a) Stretch in the spring is

L₁=sqrt(x²+L₀²) = sqrt(0.7² +0.5²)=0.86 m

Elongation is

Δx= L₁-L₀= 0.86-0.5=0.36 m.

E(total) = PE=2•k•Δx²/2=2•37.7•0.36²/2=4.9 J.

(b) PE=KE =mv²/2

v=sqrt(2•PE/m) = sqrt(2•4.9/1.37) = 2.67 m/s

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**Anonymous**, Sunday, July 20, 2014 at 2:09pm
A horizontal slingshot consists of two light, identical springs (with spring constants of 45.7 N/m) and a light cup that holds a 1.37-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.

Tries 0/4

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

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