An elastic string stretches to a total length of 30cm under a load of 500N. With an additional load of 100N, the string stretches by a further 2cm. The natural length of the string is??? Solve step by step

Answer 1

F = kx, so
∆F = k∆x
100 = 2k
k = 50 N/cm
So a load of 500N stretched the string 10 cm, making the original length 20 cm.

Answer 2

F = k x , linear stretch versus force
so how much stretch with 500 N?
(500/100) * 2 cm = 10 cm
30 cm - 10 cm = 20 cm