Answer
0.32
Work Step by Step
$F_{c}=N_{x}$
$m\frac{v^{2}}{r}=sin\theta\times\frac{mg}{cos\theta}$
$\theta=tan^{-1}\times\frac{v^{2}}{rg}$
$\theta=tan^{-1}\times\frac{\frac{65}{3.6}^{2}}{95\times9.8}$
$\theta=19.3^{\circ}$
X:$N_{y}=W+f_{y}$
$N\times cos\theta=mg+sin\theta\times kN$
$N=\frac{mg}{cos\theta-sin\theta\times k}$
y:$F_{c}=N_{x}+f_{x}$
$m\frac{v^{2}}{r}=sin\theta\times N+cos\theta\times kN$
$N=\frac{m\frac{v^{2}}{r}}{sin\theta-cos\theta\times k}$
$N=\frac{mg}{cos\theta-sin\theta\times k}=\frac{m\frac{v^{2}}{r}}{sin\theta-cos\theta\times k}$
$\frac{9.8}{cos\ 19.3^{\circ}-sin\ 19.3^{\circ}\times k}=\frac{\frac{\frac{95}{3.6}^{2}}{95}}{sin\ 19.3-cos\ 19.3\times k}$
$k=0.32$
