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A slingshot consists of a light leather cup containing a stone. The cup is pulled back against two parallel rubber bands. It takes a force of 15.0 $\mathrm{N}$ to stretch either one of these bands 1.00 $\mathrm{cm} .$ (a) What is the potential energy stored in the two bands together when a $50.0-\mathrm{g}$ stone is placed in the cup and pulled back 0.200 $\mathrm{m}$ from the equilibrium position? (b) With what speed does the stone leave the slingshot?

48.99$\frac { m } { s } \approx 49 \frac { m } { s }$

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so we have to over bands for party. We're going to say that the force, the spring constant for each rubber band would be the force of the spring divided by that stretch position. So 15 Newtons divided by 1.0 times 10 to the negative second meters or one centimeter. And this is giving us 1500 Newtons per meter. Now when the boat when both bands air stretched by 0.20 meters, this means that the total elastic potential energy So the potential energy associated with the string would be two multiplied by 1/2 k x squared. So this would simply be 1500 newtons per meter multiplied by 0.20 meters quantity squared and this is giving us 60. Jules, this would be our final answer for part A. For part B, we can apply the law of conservation of mechanical energy. Where here we simply have, um, potential energy associated with the string or associated with the rubber bands initially equaling a final kinetic energy. So we have 60 Jules equaling 1/2 mm the final squared. And to find the final, this would simply be equaling the square root of two times 60 jewels and then this would be divided by the mass of 50 times, 10 to the negative third kilograms. And this is equaling 49 meters per second. This would be our final answer for part B. That is the end of the solution. Thank you for watching.

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