Solution Found!
Figure P5.23 shows the velocity graph of a 75 kg passenger
Chapter 6, Problem 15E(choose chapter or problem)
FIGURE EX. 15 shows the velocity graph of a \(75 \mathrm{~kg}\) passenger in an elevator. What is the passenger's weight at \(t=1 \mathrm{~s}\) ? At \(5 \mathrm{~s}\) ? At \(9 \mathrm{~s}\) ?
Equation Transcription:
Text Transcription:
75 kg
t=1 s
5 s
9 s
Questions & Answers
QUESTION:
FIGURE EX. 15 shows the velocity graph of a \(75 \mathrm{~kg}\) passenger in an elevator. What is the passenger's weight at \(t=1 \mathrm{~s}\) ? At \(5 \mathrm{~s}\) ? At \(9 \mathrm{~s}\) ?
Equation Transcription:
Text Transcription:
75 kg
t=1 s
5 s
9 s
ANSWER:
Step 1 of 3
Here we have to find the apparent weights of the 75 kg passenger at the different times given.
The mass is, .
We know that weight is defined as the net normal reaction an object gets.
As the elevator is moving upwards and accelerating, the weight of the person is increasing.
The equation is as follows.
------------------(1)
Where N is the normal reaction.
G is the acceleration due to gravity.
A is the upward acceleration.