An object is hung from a spring balance attached to the ceiling of an elevator cab. the balance reads 65 n when the cab is standing still. what is the reading when the cab is moving upward (a) with a constant speed of 7.6 m/s and (b) with a speed of 7.6 m/s while decelerating at a rate of 2.4 m/s2

(A) Reading will be 65 N

(B) Net force on the elevator will be 49.076 N    

Explanation:

We have given the balance force = 65 N

Acceleration due to gravity

We know that W=mg

So

m = 6.632 kg

(a) In first case as the as the speed is constant so the force on the elevator will be 65 N

(B) In second case as the elevator is decelerating at a rate of

So net acceleration = 9.8-2.4=

So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N

answer and explanation;