Question

three masses m1 = 4.4 kg, m2 = 13.2 kg and m3 = 8.8 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above. the spring constant is 308.3
What is the distance the upper spring is extended from its unstretched length?

Answer 1

Use these equations,
$$F=kx$$
$$F=mg$$

Answer 2

that doesnt help
i have to find total unstratch length

Answer 3

m1= 4.4 kg
m2=13.2 kg
m3=8.8 kg

as BAdhi mentioned:
F=mg
g= 9.8 m/s^2 (approximately)

F1 = 4.4 * 9.8 = 43.12 N
F2 = 13.2 * 9.8=129.36 N
F3 = 8.8 * 9.8 = 86.24 N

also, as BAdhi said :

F=kx

since you need x:
x=F/k

x1 =43.12 /308.3 =0,1398637690561142 m
x2 =129.36/308.3 =0,4195913071683425 m
x3 =86.24 /308.3 = 0,2797275381122283 m

Does this help?

Answer 4

thats already given so i have to find the unstratched distance

Answer 5

I don't know what you mean by it's already given. But I think they are asking for the distances stretched from the unstretched length. They are the distances found by BoneFreeze (x1,x2,x3). Since, in F=kx, x is the lengthened distance from it's natural length