Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 150: 77

$F_s=4.9lb$

We know that $F_s=ks=14(0.5-d)~~~$eq(1) We also know that $F_s=\frac{W}{g}(\frac{v^2}{3-d})$ $\implies 14(0.5-d)=\frac{2}{32.2}[\frac{(15)^2}{3-d}]$ This simplifies to: $d=0.15ft$ Now from eq(1), we obtain: $F_s=14(0.5-0.15)$ $F_s=4.9lb$