Answer:

Step-by-step explanation:
According to hooks law "provide the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force F"

where F= applied force
k= spring constant
e= extension
Given data
length of string l = 16cm
extension e = 20-16= 4cm
applied force = 5N
we need t o first calculate the spring constant k
apply the formula


we can now calculate the extension of the string when supporting a 6N weight

The length of the string when supporting a 6N weight is

COMMENT :According to the analysis an extra weight of 1N will cause add 0.8cm to the length of the string