What is the Average Force Exerted on Occupants by Seatbelts in a Car Collision?

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SUMMARY

The average force exerted on occupants by seatbelts during a car collision can be calculated using Newton's second law, \( F = ma \). In this scenario, an 80 kg man and a 25 kg child in a 1000 kg car, traveling at 75 kph, experience forces of 8333.3 Newtons and 2083.3 Newtons, respectively, when the collision lasts for 0.2 seconds. The weight of the car does not influence the force experienced by the occupants as calculated in this context.

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  • Understanding of Newton's second law of motion
  • Basic knowledge of physics concepts related to force and acceleration
  • Ability to perform unit conversions (e.g., converting kph to m/s)
  • Familiarity with the concept of impulse and its relation to collisions
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  • Learn about the effects of different collision durations on force calculations
  • Explore real-world applications of seatbelt safety and crash dynamics
  • Investigate the role of vehicle mass in collision outcomes and occupant safety
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Homework Statement


A model of an 80 kg man and one of a 25 kg child in a 1000 kg car and secured with seatbelts. The car was moving at 75kph as it crashed into a break wall.If the collision took 0.2s, calculate the average force exerted on each model by its seat belt

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Homework Equations


f=ma


The Attempt at a Solution


im really confused, I am thinking that for the man: 80x20.8 (75000/3600), the divide by 0.2, so the force is 8333.3 Newtons?

Does the car's weight have anything to do with this?
 
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Hi Dan, welcome to PF!

:biggrin:
dan1995420 said:
im really confused, I am thinking that for the man: 80x20.8 (75000/3600), the divide by 0.2, so the force is 8333.3 Newtons?
Yes
Does the car's weight have anything to do with this?
No.
 

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