A 5.0 kg dog sits on the floor of an elevator that is accelerating downward at 1.20 m/s^2 .?

A) What is the magnitude of the normal force of the elevator floor on the dog?

i dunno if im right,my calculations show 43.05 N for both questions

i’d Say 23 center floor Up 4 = 5 familiar flooring Down 6 = 7 familiar flooring Up 10 = 11 familiar flooring; 8 Are Above the middle, One Is the middle, And 2 Are below. Then upload the three extra flooring To the precise. Now 11 flooring Are Above the middle, meaning There additionally must be 11 flooring below the middle. 11 + 11 = 22 + center = 23

Can someone explain for me to understand why the forces of question A & B are the same answer?

Clue : M = 5 kg, a = 1,20 m^2

Solution :

A) Since the elevator is accelerating downward :

W – F = M x a
(M.g) – N = M x a
(5 x 9.81) – (5 x 1.2) = N
49.05N – 6N = N
43.05 N =N

B) F = w

F = m x g
F = 5 x 3081
F = 49.05 N

A)
weight – normal force = ma
normal force = 5(9.8 – 1.2) = 43 N

B)
43 N

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