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?

Respuesta :

Answer:

784.8 Newton

1051.2 Newton

Explanation:

[tex]W=mg\\\Rightarrow W=80\times 9.81\\\Rightarrow W=784.8\ N[/tex]

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

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-10}{3}\\\Rightarrow a=-3.33\ m/s^2[/tex]

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

[tex]W=m(g-a)\\\Rightarrow W=80(9.81-(-3.33))\\\Rightarrow W=1051.2\ N[/tex]

The apparent weight of Zach would be 1051.2 Newton