Answer
(a) At point B, the speed is 16.0 m/s.
(b) At point B, the car presses on the track with a force of 11,500 N.
Work Step by Step
(a) $K_B+U_B = K_A+U_A$
$\frac{1}{2}mv^2+mg(h_B) = 0+mg(h_A)$
$v^2 = 2g~(h_A-h_B)$
$v = \sqrt{2g~(h_A-h_B)}$
$v = \sqrt{(2)(9.80~m/s^2)(25.0~m-12.0~m)}$
$v = 16.0~m/s$
At point B, the speed is 16.0 m/s.
(b) $\sum F = \frac{mv^2}{R}$
$F_N+mg = \frac{mv^2}{R}$
$F_N = \frac{mv^2}{R}- mg$
$F_N = \frac{(350~kg)(16.0~m/s)^2}{6.0~m}- (350~kg)(9.80~m/s^2)$
$F_N = 11,500~N$
At point B, the car presses on the track with a force of 11,500 N.
