A) 222 N/m
Explanation:
Hook's law gives us the relationship between force (F), displacement with respect to the original length (x) and spring's constant (k):
[tex]F=kx[/tex]
In this part of the problem, we have:
F = 120 N - 100 N = 20 N is the new force applied
x = 9.0 cm = 0.09 m is the displacement relative to the initial stretched position
Solving the equation for k, we find the spring constant
[tex]k=\frac{F}{x}=\frac{20 N}{0.09 m}=222 N/m[/tex]
B) 45 cm
We can use Hook's law also for this part of the exercise:
[tex]F=kx[/tex]
where this time we have
F = 100 N (the original pull applied)
k = 222 N/m
Solving the equation for x, we find the original displacement:
[tex]x=\frac{F}{k}=\frac{100 N}{222 N/m}=0.45 m=45 cm[/tex]
