## College Physics (4th Edition)

(a) The acceleration of the elevator is $1.8~m/s^2$ (downward) (b) The elevator's speed is $8.7~m/s$
(a) We can find the person's acceleration: $\sum F = ma$ $mg-F_N = ma$ $a = \frac{mg-F_N}{m}$ $a = \frac{(51~kg)(9.80~m/s^2)-408~N}{51~kg}$ $a = 1.8~m/s^2$ (downward) The acceleration of the elevator is $1.8~m/s^2$ (downward) (b) We can find the elevator's speed: $v_f=v_0+at$ $v_f = 1.5~m/s+(1.8~m/s^2)(4.0~s)$ $v_f = 8.7~m/s$ The elevator's speed is $8.7~m/s$