Answer
$5.4cm$
Work Step by Step
We can find the required stretch in the spring as follows:
$x=\frac{mg}{K}(\mu_scos65^{\circ}+sin 65^{\circ})$
We plug in the known values to obtain:
$x=\frac{(2.0)(9.81)}{360}(0.22cos65^{\circ}+sin65^{\circ})$
$x=0.054m$
$x=5.4cm$
