Answer
(a) $v = 38.3~m/s$
(b) The pilot's apparent weight is 3580 N.
Work Step by Step
(a) If the pilot feels weightless at the top of the loop, then the normal force $F_N$ from the plane's seat is zero and the centripetal force is provided by gravity.
At the top of the loop:
$mg = \frac{mv^2}{r}$
$v = \sqrt{gr} = \sqrt{(9.80~m/s^2)(150~m)}$
$v = 38.3~m/s$
(b) We can convert the speed to units of m/s.
$v = (280~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 77.78~m/s$
At the bottom of the loop:
$\sum F = \frac{mv^2}{r}$
$F_N -mg = \frac{mv^2}{r}$
$F_N = \frac{mv^2}{r}+mg$
$F_N = \frac{(700~N)(77.78)^2}{(9.80~m/s^2)(150~m)}+700~N$
$F_N = 3580~N$
The pilot's apparent weight is 3580 N.
