## College Physics (4th Edition)

(a) The person's acceleration is $1.40~m/s^2$ downward. (b) We can not tell if the elevator is speeding up or slowing down.
(a) Since the person's weight in the elevator is less than the person's weight on the ground, the person must be accelerating downward. Let $F_N$ be the upward force exerted on the person by the elevator. Note that $F_N = \frac{120}{140}~\times weight = \frac{6mg}{7}$. We can find the person's acceleration: $\sum F = ma$ $mg-F_N = ma$ $mg-\frac{6mg}{7} = ma$ $a = \frac{g}{7}$ $a = 1.40~m/s^2$ The person's acceleration is $1.40~m/s^2$ downward. (b) We can not tell if the elevator is speeding up or slowing down. The elevator could be moving downward and speeding up with an acceleration of $1.40~m/s^2$, or the elevator could be moving upward and slowing down at a deceleration of $1.40~m/s^2$.