A spring with a mass of 25.0 kg is stretch pass the equilibrium, point (x=0) on

Question

A spring with a mass of 25.0 kg is stretch pass the equilibrium, point (x=0) on the positive x-axis 1.50 m. The mass is released under potential energy which converts to kinetic energy a) Find the kinetic energy and potential energy for the conservative system b) Find the velocity at x-0 in a conservative system c) Find the velocity at x-o in a non-conservative system with friction, where =.350. At potential energy maximum-then released Spring Constant k=400 0000000000000001-501 In motion at x 0 Kinetic Energy? M-25.0kg At equilibrium v-? 00000000001001_-

Explanation / Answer

(A) at given instance,

PE = k x^2 /2 = 400(1.50^2)/2 = 450 J

KE = 0

(B) KE = 450 J = 25 v^2 /2

v = 6 m/s

(C) Applying work energy theorem

work done by spring + work done by friction = change in KE

450 - (0.350 x 25 x 9.8 x 1.50) = 25 v^2 /2

v = 5.07 m/s

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