## College Physics (4th Edition)

The net force on the system is $965~N$ and this net force is directed upward.
We can find the person's mass: $mg = 640~N$ $m = \frac{640~N}{9.80~m/s^2}$ $m = 65.3~kg$ Let $F_N$ be the upward force exerted on the person by the elevator. We can find the person's upward acceleration: $F_N - mg = ma$ $a = \frac{F_N-mg}{m}$ $a = \frac{700~N-640~N}{65.3~kg}$ $a = 0.919~m/s^2$ Let $M$ be the total mass of the system. We can find the net force on the system: $\sum F = Ma$ $\sum F = (1050~kg)(0.919~m/s^2)$ $\sum F = 965~N$ The net force on the system is $965~N$ and this net force is directed upward.