k 43 lbin 144 lb Solution When the cylinder dropped by 8 i

\' k = 4.3 lb/in 144 lb

Solution

When the cylinder dropped by 8 inches

the spring would stretch by 8*2 = 16 inches

k = 4.3 lb/in

work done in stretching the spring = kx²/2

                          (4.3)*(16)² /2 = 550.8 lb-in

change PE of the cylinder = 144*8 =1152 lb-in

KE of the cylinder = 1152-550.8 = 601.2 lb-in

                                      = mv²/2

mass of the cylinder = 144/32 = 4 lbm

v² = 2*601.2/4

velocity of the cylinder = 17.34 in/s

as the cylinder drops the spring will stretch more and tension in the string will increase at equilibrium the PE stored in the spring is equal to the PE change of the cylinder

let h be the drop in height of the cylinder

then PE of the cylinder = 144h

when the cylinder drops by h the spring will stretch by 2h as there are two loops of the string on the cylinder side

PE of the spring = 4.3*(2h)²/2

                           = 144 h

h = 144/8.6 = 16.74 inches

initially the spring is stretched by 4 in

the cylinder can drop maximum = 16.74 -2 = 14.74 in