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    Spring supporting two blocks without friction

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    A 0.7 lb block rests on top of a 0.5 lb block supported by but not attached to a spring of constant 9 lb/ft. The upper block is sudenly removed.
    A) Determine the maximum velocity reached by the 0.5 lb mass
    B) the maximum height reached by the 0.5 lb mass

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    From the spring formula kx=mg
    where k= spring constant =9 lb/ft
    x= displacement from normal position=x2
    mg= weight acting on the spring in this case 0.7+0.5=1.2 lb

    Solving 9 x2= 1.2 lb unstretched
    0r x2= 2/15 ft x2

    A) Determine the maximum velocity reached by the 0.5 lb mass

    Energy stored in the spring = 1/2 k x 2 = 1/2 k* (2/15) 2 = 2/25 ft-lb

    Maximum velocity is obtained when all the energy stored in the spring is converted totally to Kinetic energy
    ie 1/2 k x 2 = 1/2 m v 2

    m=W/g=0.5/32= 1/64 ...

    Solution Summary

    The solution calculates maximum velocity and maximum height.