Loose bolt falling from elevator

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SUMMARY

The discussion revolves around calculating the height from which a bolt falls when it comes loose from an elevator moving upward at 6.88 m/s. The initial equation used is d = v(initial)t + 0.5at², where the user initially calculated a distance of -48.38 meters, which was incorrect. The correct approach involves recognizing the need to separate the motion into two parts: the elevator's ascent and the bolt's subsequent fall, while also addressing sign conventions for velocity and acceleration due to gravity.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = v(initial)t + 0.5at²
  • Knowledge of gravitational acceleration, typically -9.81 m/s²
  • Familiarity with the concept of relative motion in physics
  • Ability to analyze motion in two parts (elevator ascent and bolt descent)
NEXT STEPS
  • Study the concept of relative motion in physics, focusing on how objects interact when in motion.
  • Learn about kinematic equations and their applications in real-world scenarios.
  • Investigate the effects of gravity on falling objects, including sign conventions.
  • Practice solving problems involving two-part motion scenarios, such as an object moving upward followed by a free fall.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of motion analysis in real-world applications.

jrk613
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Homework Statement



A bolt comes loose from the bottom of an elevator that is moving upward at a speed of 6.88 m/s. The bolt reaches the bottom of the elevator shaft in 3.90 s. How high up was the elevator when the bolt came loose?


Homework Equations




d= v(initial)t + .5at^2

The Attempt at a Solution


d= 6.88(3.9) + .5(-9.89)(3.9)^2

I got the answer -48.38 but the distance is wrong... anyone see where I went wrong? Thanks
 
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jrk613 said:
1.


d= v(initial)t + .5at^2

The Attempt at a Solution


d= 6.88(3.9) + .5(-9.89)(3.9)^2

I got the answer -48.38 but the distance is wrong... anyone see where I went wrong? Thanks

In the problem the displacement ang acceleration due to gravity are in the same direction, but the velocity is in the opposite direction. Accordingly use the proper signs of these quantities.
 
Changing the signs for the first half of the equation i get either + or - 26.832... for the second half i get + or - 75.21. I've tried combining all 4 combinations and all 4 are said to be wrong... any more ideas? thanks again.
 
Try this
d = - V(initial)*t + 0.5*g*t^2
 
Remember that before the bolt starts to fall it will carry on up till gravity brings it to a halt.
 
eh, no luck, thanks anyways folks... i'll find out the answer today
 
jrk613 said:
d= 6.88(3.9) + .5(-9.89)(3.9)^2

I got the answer -48.38 but the distance is wrong... anyone see where I went wrong?
Are you measuring time from when the elevator started rising or from the moment that the bolt began to fall?

If the latter, then what you have is fine except for a sign error. But perhaps the problem is the more interesting one that measures time from the moment that the elevator begins moving upward. In which case you must treat it in two parts: (1) the elevator rises to height h, and (2) the bolt falls from height h to the bottom of the shaft.
 

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