A 86-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.86 s

what does the spring scale measure in the first .86 seconds of the elevator's ascent?

acceleration= changevelocyt/time figure that out.

scale reading= mg+ma

To determine what the spring scale measures in the first 0.86 seconds of the elevator's ascent, we need to consider the forces acting on the man.

1. First, let's calculate the acceleration of the elevator using the formula:
acceleration = (final velocity - initial velocity) / time
acceleration = (1.2 m/s - 0 m/s) / 0.86 s
acceleration ≈ 1.3953 m/s²

2. Since the man is inside the elevator, the force acting on him is the normal force, which is equal to his weight.
weight = mass * acceleration due to gravity = 86 kg * 9.8 m/s²
weight ≈ 843.2 N

3. In the first stage of the elevator's ascent, the acceleration may not be constant. However, assuming it is, we can use Newton's second law of motion:
force = mass * acceleration
force = 86 kg * 1.3953 m/s²
force ≈ 120.0768 N

Therefore, the spring scale would measure approximately 120.08 Newtons in the first 0.86 seconds of the elevator's ascent.