solution;
[tex]the expression for force applied on the spring due to the load is\\
f=k\Delta x\\
here,\Delta x is the extension in the spring due to appling force\\
given three case as following\\
110N=k(40-x_{o})----------1\\
240N=k(60-x_{o})----------2\\
w=k(30-x_{o})-------------3\\[/tex][tex]To calculate the accrual length of the spring,solve th equation 1 and 2\\
\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\
0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\
0.458(60mm-x_{o})=(40mm-x_{o})\\
x_{o}(1-0.458)=(40-60(0.458))mm\\
x_{o}\frac{12.52}{0.542}\\
=23.1mm\\
to calculate the force on the spring in case,\\
solve the equation 1 and 2\\
\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\
\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\
w=\frac{110}{2.45}=44.9N[/tex]