Question

Zach, whose mass is 80 kg, is in an elevator descending at 10 m/s.The elevator takes 3.0s to brake to a stop at the first floor.

a. What is Zach's apparent weight before the elevator startsbraking ?
b. What is Zach's apparent wight while the elevator is braking?

Answer 1

784.8 Newton

1051.2 Newton

Step-by-step explanation:

`W=mg\\\Rightarrow W=80* 9.81\\\Rightarrow W=784.8\ N`

If the elevator is going down at a constant velocity then acceleration is zero and the apparent weight before braking is 784.8 N

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(0-10)/(3)\\\Rightarrow a=-3.33\ m/s^2`

While the elevator is braking it has acceleration -3.33 m/s².

`W=m(g-a)\\\Rightarrow W=80(9.81-(-3.33))\\\Rightarrow W=1051.2\ N`

The apparent weight of Zach would be 1051.2 Newton