Suppose that 2J of work is needed to stretch a spring from its natural length of 30cm to a length of 42cm

how much work is needed to stretch the spring from 35 to 40cm

So Hookies law

Work = kx^ (2)

I get

k = 2/144

The work done on the spring is equal to the elastic potential energy of a spring. We look at Hooke's law, which states that the force needed to stretch a spring is proportional to the displacement of the spring.

F = - kx

W = (integral) Fdx

W = (integral) - kx dx

W = kΔx²/2

20 = k (42 - 30) / 2

k = 10/3