Question

Answer for (B)

A spring with a spring constant of 240 N/m has a 6-kg mass attached to the end of it and is stretched a distance of .6 m. (8 points) a. What is the potential energy of the spring? b. If you let go of the spring, how fast is the mass moving when at the equilibrium point?

Answer 1

Spring constant is given, K = 240N/m. B Gravitational torce due to the maG ky mass & -mg = (6x9.8) N9-9.8 m/s = 58.8 N.. Extension of the spring due to the mass is * = som = 0. 245 m. and it is streached X2=0.6m. - to terl extension of the spring es aan,tuz = 6.2u5to.6 m =0.845m. Potential energy of the spring = KX2 =<x 210x (0.845)? = 85.683 J Potential energy due to streach is 2x Kx (0.6) = I x 240 x 0.62 = 43.25. This energy will convert to Kinetic energy at the equilibrium point. :. It v is the velocity at equilibrium point then, - mv2 – 43.2

² X 6X V² = 43.2 V = 14.4 :.V=3.79 m/s