Answer
(a) $U_{grav} = 138~J$
(b) v = 3.32 m/s
(c) Work = 0
Work Step by Step
(a) We can find the vertical height $h$ of the release point above the bottom of the swing.
$(2.20~m)~cos(42.0^{\circ}) = 2.20~m - h$
$h = 2.20~m - (2.20~m)~cos(42.0^{\circ})$
$h = 0.565~m$
We can find the potential at the release point compared to the bottom of the swing.
$U_{grav} = mgh$
$U_{grav} = (25.0~kg)(9.80~m/s^2)(0.565~m)$
$U_{grav} = 138~J$
(b) $\frac{1}{2}mv^2 = 138~J$
$v^2 = \frac{(2)(138~J)}{m}$
$v = \sqrt{\frac{(2)(138~J)}{25.0~kg}}$
$v = 3.32~m/s$
(c) Since the force of tension acts at a $90^{\circ}$ angle to the direction of motion, the tension in the ropes does zero work.
