Calculating Time of Fall: Elevator Problem

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Homework Help Overview

The problem involves an elevator ascending at a uniform velocity of 2 m/s, from which a loose bolt detaches and falls to the floor, located 3 m below. The objective is to determine the time it takes for the bolt to reach the floor, with gravitational acceleration given as 10 m/s².

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial velocity of the bolt, questioning whether it should be considered as 0 m/s or -2 m/s due to the elevator's motion. Some suggest that the problem may be analyzed from different frames of reference, such as the building or the elevator itself.

Discussion Status

There is an ongoing exploration of the implications of different reference frames on the problem. Some participants express confidence in treating the problem as if the elevator were stationary, while others question the textbook's approach and whether it leads to a different conclusion.

Contextual Notes

Participants note that the constant velocity of the elevator allows for simplifications in the analysis, but there is uncertainty regarding the textbook's reasoning and whether it results in a different answer.

Prathamesh
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Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??
 
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Perhaps the textbook is solving this from the viewpoint of motion relative to the building, i.e., relative to the building's basement?
 
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Prathamesh said:

Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??
The bolt had an upward initial velocity, while the displacement is in the downward direction.
 
Prathamesh said:

Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??
It's perfectly valid and preferable to use the frame of reference of the lift, as you did. Using the ground frame makes u=-2ms/s, but then the rest of the problem gets messier because you have to take into account the moving floor. If the textbook gets a different answer it is wrong.
 
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Prathamesh said:

Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??

Since the velocity of the elevator is constant, all laws of Physics (and of mechanics, in particular) are the same as if the elevator were not moving at all! In other words, from the point-of-view of measurements made inside the elevator, you cannot tell if it is moving or stationary. Therefore, you can neglect the elevator's motion in this question, and treat it as a simple problem of a bolt falling from ceiling to floor in your bedroom.

For more on that "not moving" analogy, see, eg.,
http://psi.phys.wits.ac.za/teaching/Connell/phys284/2005/lecture-01/lecture_01/node5.html
which concerns the so-called Galilean invariance of mechanical laws.
 
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Ray Vickson said:
Since the velocity of the elevator is constant, all laws of Physics (and of mechanics, in particular) are the same as if the elevator were not moving at all! In other words, from the point-of-view of measurements made inside the elevator, you cannot tell if it is moving or stationary. Therefore, you can neglect the elevator's motion in this question, and treat it as a simple problem of a bolt falling from ceiling to floor in your bedroom.
As I read the OP, that's what Prathamesh did. He asks why the book didn't. What is not clear is whether the book arrived at a different answer.
 
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Thank you very much haruspex and ray vickson...
I just wanted to make sure that i m going in r8 direction
 

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