Answer:
the elastic constant of the spring=1.715
the length of the spring=0.28
Step-by-step explanation:
we know that according to hooks law
F=-k x
F= force
k= elastic constant
x= extension or compression
given
length change from 35cm to 55 cm so delta x = L2-L1= 55-35=20 cm
now to find k we need F and F =ma
M for part a is 3.5 kg
so F=3.5 kg *9.8=34.3
now k=F/x
k=34.3/20=1.715 N/cm=171.5 N/m
now to find length given mass is 5 kg so
F= ma
F=5*9.8=49 N
so x =F/k
x=49/171.5
x=0.28