Answer
(a) (i) W = 0
(ii) W = 0.98 J
(b) v = 2.76 m/s
(c) The force of gravity is constant and it is directed straight down.
The normal force is not constant. The force of friction is not constant.
(d) The normal force at point B is 5.01 N
Work Step by Step
(a) (i) Since the normal force acts at a $90^{\circ}$ angle to the direction of motion, the normal force does zero work.
(ii) $W = mgR = (0.20~kg)(9.80~m/s^2)(0.50~m)$
$W = 0.98~J$
(b) $K_2 = K_1 + W$
$\frac{1}{2}mv^2 = 0+ 0.98~J-0.22~J$
$v^2 = \frac{(2)(0.76~J)}{m}$
$v = \sqrt{\frac{1.52~J}{0.20~kg}}$
$v = 2.76~m/s$
(c) The force of gravity is $mg$. It is constant and it is directed straight down.
The normal force depends on the angle of the bowl's surface. Since the angle is continuously changing, the normal force is not constant. The force of friction is also not constant because it depends on the normal force.
(d) $\sum F = \frac{mv^2}{R}$
$F_N - mg = \frac{mv^2}{R}$
$F_N = \frac{mv^2}{R} + mg$
$F_N = \frac{(0.20~kg)(2.76~m/s)^2}{0.50~m} + (0.20~kg)(9.80~m/s^2)$
$F_N = 5.01~N$
The normal force at point B is 5.01 N
