Tom has built a large slingshot, but it is not working quite right. He thinks he

Question

Tom has built a large slingshot, but it is not working quite right. He thinks he can model the slingshot like an ideal spring, with a spring constant of 55.0 N/m. When he pulls the slingshot back 0.375 m from a non-stretched position, it just doesn't launch its payload as far as he wants. His physics professor "helps" by telling him to aim for an elastic potential energy of 14.5 Joules. Tom decides he just needs elastic bands with a higher spring constant. By what factor does Tom need to increase the spring constant to hit his potential energy goal?

Explanation / Answer

The potential energy in a spring is given by the equation: U = (1/2)kx^2 For his slingshot: U = (1/2)(55.0 N/m)(0.375 m)^2 = 3.87 J To get 14.5 J 14.5 J = (1/2)k(0.375 m)^2 k = (2)(14.5J)/(0.375 m)^2 = 206.22 N/m To find the factor of increase, we divide this by the original: 206.22/55.00 = INCREASE BY A FACTOR OF 3.75 If he keeps his current spring constant: 14.5 J = (1/2)(55.0 N/m)x^2 x^2 = (2)(14.5)/(55.0) x = sqrt((2)(14.5)/(55.0)) = 0.726 m To find the factor of increase, divide by the original: 0.726/0.375 = INCREASE BY A FACTOR OF 1.94 Calculate the force when pulling the slingshot back for both scenarios: Higher k: F = (206.22 N/m)(0.375 m) = 77.33 N Higher stretch: F = (55.0 N/m)(0.726 m) = 39.93 N You have to pull harder on the increased spring constant

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