Answer
a) $\theta=35.2^o$
b) $F_T=449N$
c) $F_{Tmax}=751N$
Work Step by Step
a) $mgL(1-\cos(\theta))=\frac{mv^2}{2}$
$\cos(\theta)=1-\frac{v^2}{2gL}=1-\frac{(6.0\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(10.0m)}=0.81$
$\theta=\arccos(0.81)=35.2^o$
b) $F_T=mg\cos(\theta)=(56 kg)(9.8\frac{m}{s^2})\cos(35.2^o)=449N$
c) $F_{Tmax}=m(\frac{v^2}{L}+g)=(56kg)\Big(\frac{(6.0\frac{m}{s})^2}{10.0m}+9.8\frac{m}{s^2}\Big)=751N$
