Zach, whose mass is 80 kg, is in an elevator descending at
10 m/s. the elevator takes 3.0 s to brake to a stopat the first
floor.
a. what is zach’s apparent weight before theelevator starts
braking?
b. what is zach’s apparent weight while theelevator is braking?

a) W = 784 N

b) R = 1050.4 N

Explanation:

Given

Let us consider the upward direction positive

Mass of Zach m = 80 kg

initial velocity of the elevator u = -10 m/s

final velocity of the elevator v = 0 m/s

time taken to change velocity t = 3.0 s

Solution

a) Apparent weight before the braking

When the elevator is moving with a constant velocity, the only acceleration acts on Zach is the acceleration due to gravity

b) Apparent weight during the braking

When the elevator is breaking, its velocity changes

So its acceleration

The positive sign indicates that the acceleration is acting upward

For upward acceleration, the apparent weight

hypothesising or postulating