A 72.0 kg man stands on a spring scale in an elevator. Starting from rest, the e

Question

A 72.0 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.84 m/s in 0.500 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.80 s and comes to rest.

(a) What does the spring scale register before the elevator starts to move?
N

(b) What does it register during the first 0.500 s?
N

(c) What does it register while the elevator is traveling at constant speed?
N

(d) What does it register during the time it is slowing down?
N

Explanation / Answer

We know that,

Fnet = ma

a) Before the elevator moves,

Fnet = ma = 0 because it's not accelerating nor moving

N - mg = 0

N = mg

N = 72*9.8 = 705.6 N

b)During those 0.5 s, the elevator is accelerating to its max speed.

a = (vf - vo) / t

a = (1.84 - 0) / 0.5

= 3.68 m/s^2

Now, since it is ascending, the Normal force is going in the positive y direction, while gravity is going in the negative y direction:

N - mg = ma

N = mg + ma

N = m(g+a)

= 72*(9.8+3.68) = 970.56

c) constant speed means no acceleration.

Fnet = 0

N -mg = 0

N = mg = 72*9.8 = 705.6 N

d) since it's undergoing a negative acceleration, you want to find this acceleration:

a = (vf - vo) /t

a = (0 - 1.84)/1.8

= -1.023 m/s^2

Now plug that a into this equation:

Fnet = ma

N - mg = ma

N = m(g+a)

= 72*(9.8-1.023) = 631.94 N

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