Answer
12 m
Work Step by Step
Let's apply Newton's second law for the block through the $12^{\circ}$ slope.
$F=ma$
$F-f-mgsin12^{\circ}=ma$ ; f - frictional force
$F=f+mgsin12^{\circ}+ma$ ; Let's plug known values into this equation.
$F=68\space N+61\space kg(9.8\space m/s^{2})sin12^{\circ}+61\space kg(1.1\space m/s^{2})$
$F=259.4\space N$
This force is equal to the force exerted on the skier by the cable.
Let's apply equation 10.17 $F=Y(\frac{\Delta L}{L_{0}})A$ to find the unstrained length of the cable.
$L_{0}=\frac{YA(\Delta L)}{F}$ ; Let's plug known values into this equation.
$L_{0}=\frac{(2\times10^{11}N/m^{2})(7.8\times10^{-5}m^{2})(2\times10^{-4}m)}{259.4\space N}\approx12\space m$
The value of Y(steel) is taken from table 10.1
