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?

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

Question

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

aroub · Answer

A) w-f=ma  ( weight -normal force=ma ) 
=>f=5(9.8-1.20)
        =5(8.6)
        = 43N

aroub · Answer

B) I think it's 43N as well but I'm really not sure =( Sorry.