Answer
(a) The person's apparent weight is 784 N
(b) The person's apparent weight is 1050 N
Work Step by Step
Let $F_N$ be the normal force of the elevator floor pushing up on the person. Note that $F_N$ is equal in magnitude to the person's apparent weight.
(a) When the elevator is moving at a constant speed, $F_N$ is equal in magnitude to the gravitational force on the person.
$F_N = mg$
$F_N = (80~kg)(9.80~m/s^2)$
$F_N = 784~N$
The person's apparent weight is 784 N
(b) We can find the acceleration while the elevator is braking.
$a = \frac{0-10~m/s}{3.0~s}$
$a = -3.3~m/s^2$
The magnitude of the acceleration is $3.3~m/s^2$ and it is directed in the upward direction. We can use the acceleration to find the apparent weight.
$\sum F = ma$
$F_N-mg = ma$
$F_N = m(g+a)$
$F_N = (80~kg)(9.80~m/s^2+3.3~m/s^2)$
$F_N = 1050~N$
The person's apparent weight is 1050 N
