How far must this spring be stretched for its potential energy to be 0.020 J ?

Potential energy of spring is (1/2)k(x^2)

where k is the spring constant and x is in meters (I think...might be cm)

Also, F = kx

First solve for k.

k = F/x = 4.5/0.014 = 321.49 N/m

Now E = (1/2)k(x^2)

x = sqrt(2E/k)

x = sqrt[2(0.020)/321.49]

x = .0112 = 1.12 cm

First of all you need to find the spring constant (k) by using Hook's law: F = -kx

k = F/x = 4.5 N/0.014 m = 321.43 N/m

Then use the potential energy for a spring to find the distance: PE = 1/2kx^2

x = sqrt(2PE/k) = 0.011 m

:)