A spring of initial length 35 cm acquires a length of 55 cm when we hang from it a mass of 3.5 kg. Calculate: a) The elastic constant of the spring.
b) The length of the spring when we hang a mass of 5 Kg.

the elastic constant of the spring=1.715

the length of the spring=0.28

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

c. magnitude and direction