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A 88 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 as. It travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.4 s, and comes to rest. What does the spring scale register in each of the following time intervals?(a) before the elevator starts to move?
(b) during the first 0.80 s of the elevator 's ascent?
(c) while the elevator is traveling at constant speed?
(d) during the elevator's negative acceleration?

Respuesta :

akindelemf

Answer:

a)  Before the elevator starts to move = 862.4 N

b) During the first 0.80 s of the elevator 's ascent = 994.4 N

c) While the elevator is traveling at constant speed = 862.4 N

d) During the elevator's negative acceleration = 786.98 N

Explanation:

a)   a= 0

From Newton's second law => N - mg = 0

N = mg

= 88 x 9.8

= 862.4 N

b)

a = ΔU/Δt = 1.2/0.80 = 1.5 ms⁻²

by newton law of motion which implies N - mg = ma

N = m(g + a) = 88(9.8 + 1.5)

                    = 88 (11.3)

                    = 994.4 N

c)

For constant speed, a = 0

N = mg

   = 88 x 9.8

   = 862.4 N

d)

a = ΔV/Δt = 1.2 / 1.4 = 0.857 ms⁻²

mg - N = ma

N = m(g - a)

   = 88(9.8 - 0.857)

   = 88 (8.943)

   = 786.98 N

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