Answer
$v=9.29ft/s$, $T=38lb$
Work Step by Step
We can determine the required speed and the tension as follows:
$Wcos\theta=ma_t$
$\implies 60cos\theta=\frac{60}{32.2}a_t$
$\implies a_t=32.2cos\theta$
As $a_t=\frac{dv}{dt}=v\frac{dv}{ds}$
$\implies ds=10d\theta$
$\implies 32.2cos\theta=v\frac{dv}{10d\theta}$
$\implies 322cos\theta d\theta=vdv$
Integrating both sides, we obtain:
$\int_0^v vdv=\int_{60}^{90} 322cos\theta d\theta$
This simplifies to:
$v=9.29ft/s$
Now $\Sigma F_n=ma_n$
$\implies 2T-Wsin\theta=m\frac{v^2}{\rho}$
We plug in the known values to obtain:
$2T-(60 sin90)=\frac{60}{32.2}\times \frac{(9.288)^2}{10}$
This simplifies to:
$T=38lb$
