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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

Please see attachment

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The solution calculates maximum velocity and maximum height.

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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 ...

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