Calculating Initial Velocity: Physics of Home Alone 2's Tool Chest Scene

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SUMMARY

The discussion focuses on calculating the initial velocity of a tool chest in the movie "Home Alone 2" as it falls down the stairs after being pulled by a rope. The consensus is that the initial velocity of the tool chest is 0 m/s at the moment it begins to fall, as it starts from rest. The acceleration of the tool chest is due to gravity once it is released, leading to a subsequent increase in velocity as it descends the stairs.

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  • Basic understanding of Newton's laws of motion
  • Familiarity with concepts of acceleration and initial velocity
  • Knowledge of gravitational force and its effects on falling objects
  • Ability to apply kinematic equations in physics
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This discussion is beneficial for physics students, educators, and anyone interested in the application of physics principles in real-world scenarios, particularly in analyzing motion in film contexts.

Caronica
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Hi,

My friend and I are doing a physics project that analyzes movie physics and we have a quick question. In Home Alone 2, the kid ties a doorknob to a tool chest on wheels that is a floor up. The bad guys pull on the doorknob and the tool chest starts falling down the stairs. How would we calculate the initial velocity of the tool chest after it gets pulled by the rope?
 
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Welcome to Physics Forums.

Answer to your question: Is it not 0? The tool chest accelerates down he stairs.
 

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