#### Answer

(a) The person's apparent weight is 690 N
(b) The person's apparent weight is 740 N
(c) The person's apparent weight is 690 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) If the elevator is at rest, then the acceleration is zero. Then $F_N$ is equal in magnitude to the person's weight.
$F_N = mg$
$F_N = (60~kg)(9.80~m/s^2)$
$F_N = 690~N$
The person's apparent weight is 690 N
(b) We can find the acceleration of the elevator.
$a = \frac{v_f}{t}$
$a = \frac{10~m/s}{4.0~s}$
$a = 2.5~m/s^2$
We can use the acceleration to find the person's apparent weight.
$\sum F = ma$
$F_N- mg = ma$
$F_N = m(g+a)$
$F_N = (60~kg)(9.80~m/s^2+ 2.5~m/s^2)$
$F_N = 740~N$
The person's apparent weight is 740 N
(c) If the elevator is moving at a constant speed, then the acceleration is zero. Then $F_N$ is equal in magnitude to the person's weight.
$F_N = mg$
$F_N = (60~kg)(9.80~m/s^2)$
$F_N = 690~N$
The person's apparent weight is 690 N