Answer:
displacement (x2) = 0.09 meters
Step-by-step explanation:
let the force by the spring be Fs , force due to gravity be Fg, displacement when 2 kg mass is hung be x1 and the displacement when 6 kg mass is hung be x2.
take the upward direction as positive.
then, for both cases:
Ftot = Fs + Fg = 0 , the block gets to a complete stop.
for when 2 kg mass is hung:
Fs = Fg
k×x1 = m1×g
k = m1×g/x1 = [(2)×(9.8)]/(0.03) = 653.3 N/m
then because k remain constant since we using the same string,
for when 6 kg mass is hung:
Fs = Fg
k×x2 = m2×g
x2 = [(6)×(9.8)]/(653.3) = 0.09 meters
therefore, the string will stretch for a displacement of 0.09 meters when the 6 kg mass is hung.