## College Physics (4th Edition)

The correct answer is: (e) $896~N$
The correct answer is: (e) $896~N$ If the elevator is slowing down while descending then the elevator is accelerating upward at a rate of $3.00~m/s^2$. We can find the normal force of the scale pushing upward on the person: $F_N-mg = m~a$ $F_N = m~(g+a)$ $F_N = (70.0~kg)(9.80~m/s^2+3.00~m/s^2)$ $F_N = 896~N$