Answer
(a) v = 2.14 m/s
(b) T = 1.66 N
(c) T = 1.86 N
Work Step by Step
(a) We can find the change in vertical height $h$ from the bottom of the swing to the height when the angle is $45^{\circ}$.
$\frac{l-h}{l} = cos(45^{\circ})$
$h = l~(1 -cos(45^{\circ}))$
$h = (0.80~m)(1 -cos(45^{\circ}))$
$h = 0.234~m$
We can find the speed at the vertical position.
$\frac{1}{2}mv^2 = mgh$
$v^2 = 2gh$
$v = \sqrt{2gh}$
$v = \sqrt{(2)(9.80~m/s^2)(0.234~m)}$
$v = 2.14~m/s$
(b) $T~cos(45^{\circ}) = mg$
$T = \frac{mg}{cos(45^{\circ})}$
$T = \frac{(0.12~kg)(9.80~m/s^2)}{cos(45^{\circ})}$
$T = 1.66~N$
(c) $\sum F = \frac{mv^2}{R}$
$T - mg = \frac{mv^2}{R}$
$T = \frac{mv^2}{R} + mg$
$T = \frac{(0.12~kg)(2.14~m/s)^2}{0.80~m} + (0.12~kg)(9.80~m/s^2)$
$T = 1.86~N$
