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The 2kg collars given downward velocity of 4 m/s when it is at A. If the spring has an unstretched length of 1m and stiffness of k= 30 N/m. Determine the velocity of the collar at s=1m. Distance between collar and socket is 2m.

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Answer #1
T1 + V1 = T2 + V2 (Conservation of Mechanical Energy) T1(initial Kinetic Energy) T1=1/2(2kg)(4m/s)^2 T2(final Kinetic Energy) T2=1/2(2kg)(Vf)^2 V1(InitialGravitational and Elastic Potential Energy) V1g=2(9.81)(0) V1e=1/2(30)(2-1)^2 V2(final Gravitational and Elastic Potential Energy) V2g= -2(9.81)(1)V2e=1/2(30)((square-root of 5)-(1))^2 Substituting the values, 1/2(30)(2-1)^2 + 1/2(2)(4)^2 = 1/2(2)(Vf)^2 + 1/2(30)((square-root of 5)-(1))^2 - 2(9.81)(1)Vf= 5.26m/s (answer)
answered by: Kjd
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