A 69-kg man stands on a spring scale in an elevator. Starting from rest, the ele

Question

A 69-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 0.92 s. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.3 s, and then comes to rest.

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

(b) What does the spring scale register during the first 0.92 s of the elevator's ascent?
____N

(c) What does the spring scale register while the elevator is traveling at constant speed?
____N

(d) What does the spring scale register during the elevator's negative acceleration?
____N

Explanation / Answer

Given ,

weight of man = 69kgs

the elevator attains a maximum speed of 1.2m/sec in 0.92 sec

the elevator travels with a constant speed for 5 sec and undergoes a uniform negative acceleration for 1.3 sec and then comes to rest.

a) Fnet = ma where the two forces are Normal force(spring scale reading) and weight of man (mg)

before the elevator moves Fnet = ma =0 because it is not accelerating nor moving

N-mg =0

N = mg

N = 69*9.8 = 676.2 N

b) during those 0.92 sec the elevator is ascending towards its maximum speed

a= (Vf - Vo)/t = (1.2 - 0)/ 0.92

= 1.304 m/sec^2

Now Since it is ascending the Normal force is moving in Positive Y direction while the Gravity is in Negative Y direction.

N-mg = ma

N = m(g+a) = 69(9.8+1.304) = 766.2 N

c) Constant speed means No acceleration

Fnet = 0

N - mg =0

N=mg = 69*9.8

= 676.2 N

d) Since it is undergoin a negative acceleration

a = (Vf-Vo)/t= (0 - 1.2)/1.3

a = -0.92 m/sec^2

Fnet = ma

N - mg = ma

N = m(a+g) = 69(-0.92+9.8) = 612.72 N

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