A mass of 2.0 kg stretches a verticle spring 15 cm and the mass comes to rest in

Question

A mass of 2.0 kg stretches a verticle spring 15 cm and the mass comes to rest in a new equilibrium position. If a someone now pulls the mass down and stretches the spring an additional 8 cm and releases, calculate the max speed of the mass. I was just wondering if my answer is correct. I have .83m/s A mass of 2.0 kg stretches a verticle spring 15 cm and the mass comes to rest in a new equilibrium position. If a someone now pulls the mass down and stretches the spring an additional 8 cm and releases, calculate the max speed of the mass. I was just wondering if my answer is correct. I have .83m/s

Explanation / Answer

The spring constant of the spring is k = F/x = (2.0 kg)(9.80 m/s2)/(0.15 m) = 130.66 N/m Given that the spring is streched additional 8.0 cm So x = 0.15 m+ 0.08 m = 0.23 m We have k = m1g/x The mass m1 = kx/g = (130.66 N/m)(0.23 m)/(9.80 m/s2) = 3.06651 kg From the law of conservation of energy (1/2)kx^2 = (1/2) mv^2 (130.66 N/m)(0.15 m)^2 = (2.0 kg)v^2 The maximum speed of the mass is v= 1.2124 m/s

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