Pearson eText for College Physics: A Strategic Approach -- Instant Access (Pearson+)
Pearson eText for College Physics: A Strategic Approach -- Instant Access (Pearson+)
4th Edition
ISBN: 9780137561520
Author: Randall Knight, Brian Jones
Publisher: PEARSON+
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Chapter 5, Problem 19P

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10m/s. A 60 kg passenger gets aboard on the ground floor. What is the passenger's apparent weight

a. Before the elevator starts moving?

b. While the elevator is speeding up?

c. After the elevator reaches its cruising speed?

a.

Expert Solution
Check Mark
To determine

To find: The apparent weight of the passenger before the elevator starts moving.

Answer to Problem 19P

Solution: The apparent weight of the passenger before the elevator starts moving is 590N.

Explanation of Solution

Given data:

Mass of the passenger (m) is 60 kg.

Cruising speed of the elevator is 10ms.

Time taken by the elevator to reach its cruising speed is 4.0 s.

Formula used:

The expression for apparent weight of the passenger before the elevator starts moving is as follows:

wapp=m(g+ay) (1)

Here,

m is the mass of the passenger.

g is the acceleration due to gravity, which is 9.8ms2, and

ay is the acceleration of the passenger.

Two forces act on the passenger along a single vertical line. They are the downward pull of the gravity and the upward push of the elevator.

The elevator is at rest and so is the passenger before the elevator starts moving up. Therefore, the acceleration of the passenger before the elevator starts moving is 0ms2.

ay=0ms2

Calculation of apparent weight of the passenger before the elevator starts moving:

Substitute 60 kg for m, 9.8ms2 for g, and 0ms2 for ay in Equation (1).

wapp=(60kg)(9.8ms2+0ms2)=(60kg)(9.8ms2)=588kgms2590N {1kgms2=1N}

Conclusion:

Thus, the apparent weight of the passenger before the elevator starts moving is 590N.

b.

Expert Solution
Check Mark
To determine

To find: The apparent weight of the passenger while the elevator is speeding up.

Answer to Problem 19P

Solution: The apparent weight of the passenger while the elevator is speeding up is 740N.

Explanation of Solution

The expression to find the acceleration is as follows:

ay=vfviΔt (2)

Here,

vi is the initial speed of the elevator.

vf is the final speed of the elevator (cruising speed), and

Δt is the time taken by the elevator to reach its cruising speed.

The speed of the elevator initially is 0ms and gradually reaches its cruising speed of 10ms in 4 seconds.

Therefore, the initial speed of the elevator is 0ms.

vi=0ms

And, the final speed of the elevator is 10ms.

vf=10ms

Calculation of acceleration:

Substitute 0ms for vi, 10ms for vf, and 4.0 s for Δt in Equation (2).

ay=10ms0ms4.0s=10ms4.0s=2.5ms2

Calculation of apparent weight of the passenger, while the elevator is speeding up:

Substitute 60 kg for m, 9.8ms2 for g, and 2.5ms2 for ay in Equation (1).

wapp=(60kg)(9.8ms2+2.5ms2)=(60kg)(12.3ms2)=738kgms2740N

Conclusion:

Thus, the apparent weight of the passenger while the elevator is speeding up is 740N.

c.

Expert Solution
Check Mark
To determine

To find: The apparent weight of the passenger after the elevator reaches its cruising speed.

Answer to Problem 19P

Solution:

The apparent weight of the passenger after the elevator reaches its cruising speed is 590N.

Explanation of Solution

As the elevator does not accelerate anymore after it reaches its cruising speed, the acceleration of the passenger is 0ms2.

ay=0ms2

Calculation of apparent weight of the passenger after the elevator reaches its cruising speed:

Substitute 60 kg for m, 9.8ms2 for g, and 0ms2 for ay in Equation (1).

wapp=(60kg)(9.8ms2+0ms2)=(60kg)(9.8ms2)=588kgms2590N

Conclusion:

Thus, the apparent weight of the passenger after the elevator reaches its cruising speed is 590N.

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Chapter 5 Solutions

Pearson eText for College Physics: A Strategic Approach -- Instant Access (Pearson+)

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