Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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