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Raza
Dec4-04, 12:40 AM
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

so-crates
Dec4-04, 01:11 AM
a) Look in your textbook or http://www.wikipedia.org for the definition of KE

futb0l
Dec4-04, 05:45 AM
Look for the Work Energy Theorem or Conservation of Mechanical Energy...

prasanna
Dec4-04, 07:37 AM
Kinetic Energy is the energy possessed by a body due to its velocity
KE = \frac{1}{2}mv^2

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

Try to find them out!! o:)

Raza
Dec4-04, 10:31 AM
I know momentum is M=mv but I never heard of impulse.
so KE would be:
1/2(75)(15)^2
:632812.5

Raza
Dec4-04, 05:30 PM
Did I do it right?

Raza
Dec4-04, 09:49 PM
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 :)

futb0l
Dec5-04, 12:16 AM
Yeah - u got the KE right :)

futb0l
Dec5-04, 12:19 AM
for b ..

I think it is...


W = \sum F \Delta x = KE_f - KE_i


the final kinetic energy is 0 - cuz it comes to rest...
the displacement is 0.50m - so then you should be able to solve for F.

Raza
Dec5-04, 09:48 AM
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.

HallsofIvy
Dec5-04, 10:22 AM
If I were your teacher I would NOT say that
:632812.5
1/2(75)(15)^2

What are the units?

Raza
Dec5-04, 11:32 AM
I noticed I did A wrong. It's suppose to like this:

A) (1/2)(75)(15^2)

(1/2)(16875)

8437.5J/1000

:8.43kJ

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

\frac{Ke}{D} = \frac{A}{1m}

Given:
Ke:8.4kJ
D:0.5m

\frac{8.4kJ}{0.5m} = \frac{A}{1m}

Now, we can solve for A

A=\frac{(8.4kJ)(1m)}{0.5m}

A=16.8kN

Nylex
Dec5-04, 11:50 AM
[b]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.