Two identical springs (neglect their masses) are used to Solution I will assume

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Two identical springs (neglect their masses) are used to

Explanation / Answer

I will assume that they mean m = 160g = .160kg . Conservation of energy tells us that the initial potential energy of the compressed lower spring --- Pe = 1/2kx^2 = 1/2k(.1)^2= .005k is converted completely to kinetic energy at the equilibrium position of the lower spring , and this is the total energy of the mass : E= Pe = .005k = Ke = 1/2(m)Vo^2= 1/2(.16)Vo^2= .08Vo^2 (1) The total energy of the mass is then in general : E = .005k = mgh + 1/2mV^2 + 1/2kx^2 (2) where 1/2kx^2 is the potential stored in the upper spring when we start commpressing it . At the equilibrium position of the upper spring , the upper spring has not been compressed yet ; so , x =0 . We can then write using equation (2) at the equilibrium position of the upper spring for the mass : E= .005k = (.16)(9.8)(.3) + 1/2(.16)Vb^2 (3) At the maximum position of the mass , h = .3+.02 = .32 , V =0 , and x = .02 . We can substitute these values into equation (2) to solve for k : E = .005k = (.16)(9.8)(.32) + 1/2(.16)(0)^2 + 1/2k(.02)^2 .005k = .50176 + .0002k . We can solve this equation for k : (.005-.0002)k = .50176 .0048k = .50176 -----> ------------------------------------------------------------------------------------------------------------ k = (.50176)/(.0048) = 104.53 Newtons/meter -------- solution ------------------------------------------------------------------------------------------------------------- You could then use this value of k to find the velocity at the two equilibrium positions of the mass on the springs when the mass is at either of these positions with equations (1) or (3) .

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