What average force did the seat belt exert on the passenger during the crash?

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Homework Help Overview

The discussion revolves around a physics problem involving a passenger in a van during a collision. The problem requires calculating the passenger's kinetic energy before the crash and determining the average force exerted by the seat belt during the crash. The subject area includes concepts of kinetic energy, momentum, and impulse.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of kinetic energy and its calculation, with some providing the formula. Questions arise regarding the concepts of momentum and impulse, and how they relate to the problem. There is also a request for clarification on the calculations and units involved.

Discussion Status

Some participants have confirmed the calculation of kinetic energy, while others are exploring the relationship between work, force, and displacement in the context of the second part of the problem. There is a mix of understanding and confusion regarding the concepts, with some guidance provided on the relevant equations.

Contextual Notes

Participants express a need for clarity on definitions and calculations, indicating that this problem is part of a larger unit test and assignment. There is mention of potential errors in calculations and a focus on understanding the underlying physics concepts.

Raza
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Alright, so here's the problem:

A 75kg passenger in a van is wearing a seat belt when the van, moving at 15m/s, collides with the concrete wall. The front end of the van collapses 0.50m as ot comes to rest.

a)What was the passenger's kinetic energy before the crash?

b)What average force did the seat belt exert on the passenger during the crash?

I would show my work but I don't even know where to start.
And could anyone please explain to me briefly what is kinetic energy?

Thanks
 
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Look for the Work Energy Theorem or Conservation of Mechanical Energy...
 
Kinetic Energy is the energy possessed by a body due to its velocity
KE = [tex]\frac{1}{2}mv^2[/tex]

However, for this question, you will need somethind called momentum and impulse .

Try to find them out! o:)
 
I know momentum is M=mv but I never heard of impulse.
so KE would be:

[tex]1/2(75)(15)^2[/tex]
[tex]:632812.5[/tex]
 
Did I do it right?
 
Could someone please respond? I have to understand this for my unit test and I also need to hand in a assignment regarding this question.
Thanks :)
 
Yeah - u got the KE right :)
 
for b ..

I think it is...

[tex] W = \sum F \Delta x = KE_f - KE_i[/tex]

the final kinetic energy is 0 - because it comes to rest...
the displacement is 0.50m - so then you should be able to solve for F.
 
  • #10
Could you explain it to me more about question B?
I don't even know what the big, E-looking thing is or the triangle.
 
  • #11
If I were your teacher I would NOT say that
[tex]:632812.5[/tex]
[tex]1/2(75)(15)^2[/tex]

What are the units?
 
  • #12
I noticed I did A wrong. It's suppose to like this:

A) [tex](1/2)(75)(15^2)[/tex]

[tex](1/2)(16875)[/tex]

[tex]8437.5J/1000[/tex]

[tex]:8.43kJ[/tex]

And I also figured out B, thanks futb0l!
This is what I did:

[tex]\frac{Ke}{D} = \frac{A}{1m}[/tex]

Given:
Ke:8.4kJ
D:0.5m

[tex]\frac{8.4kJ}{0.5m} = \frac{A}{1m}[/tex]

Now, we can solve for A

[tex]A=\frac{(8.4kJ)(1m)}{0.5m}[/tex]

[tex]A=16.8kN[/tex]
 
Last edited:
  • #13
Raza said:
I know momentum is M=mv but I never heard of impulse.


Usually p is used for momentum. Impulse is defined as I = Ft = change in momentum.
 

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