Question

Zach, whose mass is 80 kg, is in an elevator descending at

Answer 1

(a) apparent weight is actual weight = mg = (79)(9.81) = 775 N ANS (a) {reason: elevator is not accelerating therefore Z's net acceleration is "g"} 
(b) elevator decelerates at 12/2.5 = -4.80 m/s this reduces net acceleration acting on Z to (9.81 - 4.80) = 5.01 m/s² while braking with 5.01 m/s² acting on Z, his apparent weight = (79)(5.01) = 396 N ANS (b)

Answer 2

Answer:

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

$W = mg\\\\W = 80 \times 9.8 \\\\W = 784 N$

b) Apparent weight during the braking

When the elevator is breaking, its velocity changes

So its acceleration

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{0-(-10)}{3} \\\\a = 3.33 m/s^2$

The positive sign indicates that the acceleration is acting upward

For upward acceleration, the apparent weight

$R = m(g+a)\\\\R = 80 ( 9.8 + 3.33)\\\\R = 1050.4 N$