Answer
At the top of the path, the magnitude of the normal force is 0.456 N.
Work Step by Step
Let point A be the bottom of the loop. We can find the speed at point A.
$\frac{mv_A^2}{R} = \sum F$
$\frac{mv_A^2}{R} = F_N-mg$
$v_A^2 = \frac{R~(F_N-mg)}{m}$
$v_A = \sqrt{\frac{R~(F_N-mg)}{m}}$
$v_A = \sqrt{\frac{(0.800~m)[3.40~N-(0.0500~kg)(9.80~m/s^2)]}{0.0500~kg}}$
$v_A = 6.82~m/s$
Let point B be the top of the loop. We can find the speed at point B.
$K_B+U_B = K_A+U_A$
$K_B = K_A+0 - U_B$
$\frac{1}{2}mv_B^2 = \frac{1}{2}mv_A^2-2mgR$
$v_B^2 = v_A^2-4gR$
$v_B = \sqrt{v_A^2-4gR}$
$v_B = \sqrt{(6.82~m/s)^2-(4)(9.80~m/s^2)(0.800~m)}$
$v_B = 3.89~m/s$
We can find the normal force $F_N$ at point B.
$\sum F = \frac{mv_B^2}{R}$
$F_N+mg = \frac{mv_B^2}{R}$
$F_N = \frac{mv_B^2}{R}-mg$
$F_N = \frac{(0.0500~kg)(3.89~m/s)^2}{0.800~m}-(0.0500~kg)(9.80~m/s^2)$
$F_N = 0.456~N$
At the top of the path, the magnitude of the normal force is 0.456 N.
