If 5.4 J of work are needed to stretch a spring from 12 cm to 18 cm and 9 J are
ID: 3039600 • Letter: I
Question
If 5.4 J of work are needed to stretch a spring from 12 cm to 18 cm and 9 J are needed to stretch it from 18 cm to 24 cm, what is the natural length of the spring?Explanation / Answer
knowing the function ==> f (x) = kx we know if we integrate : ? kx = W calling the natural length as L, then the length is as 0.08 - L 0.1-L ? kx = W =====> and k is constant 0.08-L . . . .. . . . .0.1-L (1/2) k x^2 ] = 9 . . . .. ... . 0.08-L k [ (0.1-L)^2 - (0.08-L)^2 ] = 18 k [ 0.0036 - 0.04L ] = 18 ====> k = [ 1 8 ] / [ 0.0036 - 0.04L ] to the second work: 0.12-L ? kx = 15 =====> and k is constant 0.1-L . . . .. . . . .0.12-L (1/2) k x^2 ] = 15 . . . .. ... . 0.1-L k [ (0.12-L)^2 - (0.1-L)^2 ] = 30 k [ L^2 - 0.24L + 0.0144 - (L^2 - 0.2L + 0.01) ] = 30 k [ 0.0044 - 0.04L ] = 30 =====> k = [ 30 ] / [ 0.0044 - 0.04L ] k = k [ 18 ] / [ 0.0036 - 0.04L ] = [ 30 ] / [ 0.0044 - 0.04L ] 18 * ( 0.0044 - 0.04L ) = 30 * ( 0.0036 - 0.04L ) 0.0792 - 0.72L = 0.108 - 1.2 L 0.0792 - 0.108 = -1.2 L + 0.72L -0.0288 = -.48 L L = 0.06 m which is 6 cm