A 78-kg man stands on a spring scale in an elevator. Starting from rest, the ele

Question

A 78-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 0.90 s. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.2 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 0.90 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

According to the kinematic equations

V=u+ a_x t

Starting from rest, the spring scale acceleration is

a_x=(V-u)/(t)

   =(1.2-0)/0.90

   = 1.33 m/s^2

The elevator travels with this constant speed for undergoes a uniform negative acceleration   

V=u+a_y t

a_y=V-u/t

a_y=0-1.2/1.2

a_y=-1 m/s^2

a) from newtons 2nd law

F=ma

the spring scale register before the elevator starts to move

F=78(9.8)

=764.4 N

b)

the spring scale register during the first of the elevator's ascent

F=m(g+a_x)

F=78(9.8+1.33)

F=868.14 N

c)

the spring scale register while the elevator is traveling at constant speed

F=mg

=78(9.8)

=764.4 N

d)

the spring scale register during the elevator's negative acceleration

F=m(g-a_y)

=78(9.8-(-1))

=78(10.8)

=842.4 N

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