A 60.0-kg person rides in elevator while standing on a scale. The elevator has an acceleration 2.00 m/s² upward. What is the reading on the scale? A) 589 N B) 709 N C) 469 N D) zero E) 349 N

Question

Quizplus · Accepted Answer

We can use Newton's second law, F=ma, where F is the net force acting on the person, m is the person's mass, and a is the acceleration of the elevator. In this case, the net force is the difference between the force of gravity pulling the person down and the normal force of the scale pushing the person up. So, we have:
Fnet = ma
Fgravity - Fscale = ma
mg - Fscale = ma
Fscale = mg - ma
Plugging in the given values, we get:
Fscale = (60.0 kg)(9.81 m/s^2) - (60.0 kg)(2.00 m/s^2)
Fscale = 588.6 N - 120 N
Fscale = 468.6 N
Therefore, the reading on the scale is 468.6 N, which is closest to choice B, 709 N.

A 600-Kg Person Rides in Elevator While Standing on a Scale