A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator later stops, the scale reading is 400 N. Assume the magnitude of the acceleration is the same during starting and stopping. (a) Determine the weight of the person. N (b) Determine the person's mass. kg (c) Determine the magnitude of acceleration of the elevator. m/s2

a) 496Nb) 50.56kgc) 1.90m/s²

Explanation:

According to newton's secomd law, ∑F = ma

∑F is the summation of the force acting on the body

m is the mass of the body

a is the acceleration

Given the normal force when the elevator starts N1 = 592N

Normal force after the elevator stopped N2 = 400N

When the elevator starts, its moves upward, the sum of force ∑F = Normal (N) force on the elevator - weight of the person (Fg)

When moving up;

N1 - Fg = ma

N1 = ma + Fg ... (1)

Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;

N2 - Fg = - ma

N2 = - ma+Fg ... (2)

Adding equation 1 and 2 we will have;

N1+N2 = 2Fg

592N + 400N = 2Fg

992N 2Fg

Fg = 992/2

Fg = 496N

The weight of the person is 496N

/b) To get the person mass, we will use the relationship Fg = mg

g = 9.81m/s

496 = 9.81m

mass m = 496/9.81

mass = 50.56kg

c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;

N1-N2 = 2ma

592-400 = 2 (50.56) a

192 = 101.12a

a = 192/101.12

a = 1.90m/s²