A 60.0-kg person rides in an elevator while standing on a scale.The scale reads .The acceleration of the elevator is closest to A) 3.13 m/s² downward. B) 6.67 m/s² downward. C) zero. D) 9.80 m/s² downward. E) 6.67 m/s² upward.

Question

Quizplus · Accepted Answer

The scale reading gives us the apparent weight of the person which includes the weight due to gravity and the weight due to the elevator's acceleration. We can use Newton's second law to find the acceleration of the elevator:
F_net = m*a
where F_net is the net force on the person (which is equal to the apparent weight), m is the person's mass, and a is the elevator's acceleration.
F_net = m*g + m*a
where g is the acceleration due to gravity.
Solving for a:
a = (F_net - m*g)/m
a = (11ea83ea_6973_bd97_aeff_b76dcc1989f3_TB6394_11 - 60*9.8)/60
a = -3.13 m/s^2 (downward)
Therefore, the acceleration of the elevator is 3.13 m/s^2 downward (choice A).

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