Question

A spring whose natural length is 1010 cm exerts a force of 3030 n when stretched to a length of 1515 cm. (note 1515 cm - 1010 cm = 55 cm = .05.05 m ) find the spring constant k=k= 2 determine the work done in stretching the spring 77 cm ( =.07=.07 m) beyond its natural length. w=w= joules

Answer 1

The equation that relates the spring extension to the force applied to it is linear:
$F=k*\Delta x$
where $k$ is the spring constant. 
Hence $k= \frac{F}{\Delta x}= \frac{3030}{(15.15-10.10)}= \frac{3030}{5.05}=600 ( \frac{N}{m} )$
because $1 m = 100 cm$
The work done when stretching the spring is
$W= -\int\limits^d_0 {F(x)} \, dx= -\int\limits^d_0 {(kx)} \, dx = -\frac{kx^2}{2}= \frac{600*0.77^2}{2}= -177.87 (Joules)$
The work is negative because we stretch the spring while the string oposes to this stretching.