The elastic potential energy stored by a spring is
[tex]U= \frac{1}{2} k x^2 [/tex]
where [tex]k=1.1 N/m[/tex] is the spring's constant and x is the displacement of the spring with respect its rest position.
In the problem we have two situations: situation 1 (initial condition), where the spring has a displacement of [tex]x_1 = -3.9 m[/tex], and situation 2 (final condition), where the spring has a displacement of [tex]x_2 = -1.5 m[/tex]. The work performed by the spring corresponds to its loss in elastic potential energy:
[tex]W=U_1- U_2 = \frac{1}{2}kx_1^2- \frac{1}{2}kx_2^2 = \frac{1}{2}k[x_1^2-x_2^2]= [/tex]
[tex]= \frac{1}{2}(1.1 N/m)[(-3.9m)^2-(-1.5m)^2]=71 J [/tex]
this work has positive sign, because it is performed by the spring (in factm the spring is releasing from a position of higher potential energy to a position with less potential energy).
Therefore, the correct answer is The spring performs 71 J of work.
