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

In summary: So, for this question, you would use p instead of M.\In summary, the passenger's kinetic energy before the crash was 632812.5 Joules and the average force the seat belt exerted on the passenger during the crash was 8437.5 Joules.
  • #1
Raza
203
0
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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  • #3
Look for the Work Energy Theorem or Conservation of Mechanical Energy...
 
  • #4
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:)
 
  • #5
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]
 
  • #6
Did I do it right?
 
  • #7
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 :)
 
  • #8
Yeah - u got the KE right :)
 
  • #9
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.
 

1. What is the definition of average force?

The average force is the total force exerted over a period of time divided by the duration of that time. In other words, it is the average amount of force applied to an object.

2. How is the average force calculated?

The average force can be calculated by dividing the change in momentum of an object by the time interval in which the change occurred. It can also be calculated by dividing the work done on an object by the distance over which the force is applied.

3. Why is it important to know the average force exerted by a seat belt during a crash?

The average force exerted by a seat belt during a crash is important because it can help determine the level of injury sustained by the passenger. If the force is too high, it may cause serious injuries, but if it is too low, the passenger may not be properly restrained and can still be at risk of injury.

4. How does the average force exerted by a seat belt vary from person to person?

The average force exerted by a seat belt can vary from person to person based on several factors such as body weight, height, and position in the vehicle. It can also vary based on the type and condition of the seat belt and the speed and severity of the crash.

5. Can the average force exerted by a seat belt be accurately determined?

Yes, the average force exerted by a seat belt can be accurately determined through crash simulations and experiments. These methods use advanced technologies and data analysis to calculate the force exerted on a passenger during a crash.

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