it is determined that 600 J of work are necessary to stretch

Solution

l be length of spring

k be spring constant

work done to stretch spring to 10 cm(0.1m)

W = (1/2)k(0.1-l)²

work done to stretch spring to 12 cm (0.12 m)

W+6 = (1/2)k(0.12-l)²

work done to stretch spring to 14 cm (0.14 m)

W+6+10 = = W+16=(1/2)k(0.14-l)²

6=(1/2)k[(0.12-l)²-(0.1-l)²] = (1/2)k(0.0144-0.24l+l²-0.01+0.2l-l²) = (1/2)k(0.0044-0.04l)

similarly

10=(1/2)k[(0.14-l)²-(0.12-l)²] = (1/2)k(0.0196-0.28l+l²-0.0144+0.24l-l²) = (1/2)k(0.0052-0.04l)

similarly

16=(1/2)k[(0.14-l)²-(0.1-l)²] = (1/2)k(0.0196-0.28l+l²-0.01+0.2l-l²) = (1/2)k(0.0096-0.08l)

solving htese equations

10 - 6 =[(1/2)k(0.0052-0.04l)] - [(1/2)k(0.0044-0.04l)]

4 = (1/2)k(0.0008)

k = 10000

spring constant = 10000 N/m

16 - 6 = [(1/2)k(0.0096-0.08l)] - [(1/2)k(0.0044-0.04l)]

10 = (1/2)k[0.0052-0.04l] = (1/2)1000[0.0052-0.04l]=(1/2)(52-400l) = 26-200l

200l = 16

l = 0.08m

length of spring = 8cm