Suppose that 5 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 39 cm.(a) How much work is needed to stretch the spring from 32 cm to 34 cm? (Round your answer to two decimal places.)(b) How far beyond its natural length will a force of 25 N keep the spring stretched? (Round your answer one decimal place.)

Answergiven,work = 5 Jspring stretch form 30 cm to 39 cm[tex]W = \dfrac{1}{2}kx^2[/tex]x = 0.39 - 0.30 = 0.09 m[tex]5 = \dfrac{1}{2}k\times 0.09^2[/tex][tex]k = \dfrac{5\times 2}{0.09^2}[/tex]k = 1234.568 N/ma) work when spring is stretched from  32 cm to 34 cmx₂= 0.34 -0.30 = 0.04 mx₁ = 0.32 - 0.30 = 0.02[tex]W = \dfrac{1}{2}k(x_2^2-x_1^2)^2[/tex][tex]W = \dfrac{1}{2}\times 1234.568 \times (0.04^2-0.02^2)^2[/tex]W = 0.741 Jb) F = k x   25 = 1234.568 × x      x = 0.0205 m      x = 2.05 cm