Answer
$k=9.88KN/m$
Work Step by Step
We can determine the stiffness of the spring as follows:
$F_s=k[(600sin\theta-600sin15)-(200sin\theta-200sin15)]$
$\implies F_s=400k(sin\theta-sin15)$
The virtual displacements are given as
$\delta_{yA}=\frac{d(200sin\theta)}{d\theta}=200cos\theta$
$\delta_{yB}=\frac{d(600sin\theta)}{d\theta}=600cos\theta$
and $\delta_{yC}=\frac{d(1600sin\theta)}{d\theta}=1600cos\theta$
Now the virtual-work equation is
$\delta U=0$
$\implies P\delta_{yC}+F_s(\delta_{yB}-\delta_{yA})=0$
We plug in the known values to obtian:
$600(1600cos\theta)-400k(sin\theta-sin15)(600cos\theta-200cos\theta)=0$
This simplifies to:
$k=9.88KN/m$
