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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

Verified Answer :

MissPhiladelphia

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

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