Question

Consider a 77.0-kg man standing on a spring scale in an elevator. Starting from rest, the...

Consider a 77.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.08 m/s in 0.900 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 1.70 s and 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 0.900 s?
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 time it is slowing down?
N
0 0
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Answer #1
Concepts and reason

Newton’s second law:

According to which, the force applied to a body in a direction is directly proportional to the mass and acceleration of the body along that direction.

Newton’s second law can be expressed as follows:

F =ma

Here, applied force is F, mass of the object is m and the acceleration of object is a.

Spring scale also termed as newton meter refers to a typical weighing scale that measures the force like weight of an object from the deflection made in the spring in which the object is held. Its working is based on Hooke’s law; the force used to stretch or compress an elastic body is in proportion with the deflection caused in the elastic body.

Spring scale’s reading refers to the weight of the man which is the calculated as his mass times the acceleration. Rate of change in velocity of elevator is the acceleration of the man. From the Newton’s second law, force due to acceleration can be computed as spring scale reading can be computed.

Fundamentals

Acceleration:

Change in velocity with respect to time of an object is called acceleration of that object. Its S.I unit is m/s
.

Write the formula to calculate acceleration(a)
.

-u
t

Here, final velocity is V
, initial velocity is и
and the time taken is t
.

Write the formula for weight of an object.

W mg

Here, the weight of the object is W, the mass of the object is m, and the acceleration due to gravity is g.

(a)

Calculate the force due to weight of the man (w)
.

W mg

Here, the mass of man is т
and the acceleration due to gravity is .

Substitute 77 kg
for т
and 9.81 m/s2
for .

1N
W =(77 kg)(9.81 m/s* )*1 kg +m/s?
=755.37 N

(b)

Calculate the acceleration of elevator(a)
during first 0.900 s
.

-u
t

Here, final velocity is V
, initial velocity is и
and the time taken is t
.

Substitute 1.08 m/s
for V
, for и
and 0.900 s
for t
.

1.08 m/s-0
a=
0.900 s
1.2 m/s

Calculate the force that the spring scale registers during first 0.900 s
(F)
.

FW

Here, force due to acceleration of elevator is F
a
.

Substitute та
for F
a
from Newton’ second law.

F 3D та+ W

Substitute for т
, 1.2 m/s2
for and 755.37 N
for W
.

1N
(77 kg) 1.2 m/s)x-ko.m/s+755.37 N
847.77 N

(c)

Calculate the force that the spring scale registers(F,)
when the elevator is moving at constant speed.

F(F)W

Here, force due to acceleration of elevator is (F.)
.

From Newton’ second law, plug in та,
for (F.)
.

Е, %3 та, + W

Substitute for , for а,
and for .

F 0+755.37N
F 755.37 N

(d)

Calculate the deceleration of elevator(a,)
during slowing down.

а, в-

Here, final velocity after elevator slows down is V2
and the time taken to slow down is .

Substitute for , for V2
and 1.7 s
for .

0-1.08 m/s
а,
1.7s
-0.635 m/s2
а,

Calculate the force that the spring scale registers(F)
when the elevator is slowing down.

F(F)W

Here, force due to deceleration of elevator is (F)
.

From Newton’ second law, plug in та,
for (F)
.

F 3 та, + W

Substitute for , -0.635 m/s
for а,
and for .

N
F(77 kg)(-0.635 m/sJ*1 kg-m/s +755.37 N|
F 706.475 N

Ans: Part a

The spring scale registers a weight (w)
of 755.37 N
before the elevator starts moving.

Part b

The spring scale registers a force (F)
of 847.77 N
for the first 0.900 s
.

Part c

The spring scale registers a force (F,)
of when the elevator is moving at constant speed.

Part d

The spring scale registers a force (F)
of 706.475 N
when the elevator is slowing down.

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