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Force and newton's laws (airbag)

  • Thread starter missrikku
  • Start date
  • #1
missrikku
Hi, we're starting our chapter on Newton's Laws and force and I wanted to know if I am approaching the following problem correctly:

A car travels 53 km/h and hits a bridge abutment (what's that?). The person in the car moves forward 65 cm (w/respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 41 kg?

Well, I believe that the Vo of the car will also be the Vo of the person. So:

Voc = Vop = Vo = 53 km/h = 14.72 m/s

The distannce the person moved forward:

Dp = X-Xo = 65 cm = 0.65 m

Because the airbag brought the person to rest:

Vf = 0 m/s

We now have Vo, X-Xo, and Vf. We can find a:

V^2 = Vo^2 + 2a(X-Xo) --> a = -166.676 m/s^2

Using F=ma:

F = ma = (41)(-166.676) = -6833.7 N

Because they are looking for magnitude, I can take the abs value of F and get my answer: 6.8 x 10^3 N

Was my approach correct? Is F negative because a was negative, meaning it was decelerating?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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A "bridge abutment" is the pillar on one side of the road holding the bridge up. Not at all a nice thing to hit at 53 km/hr- they are large, heavy, and have a bad attitude.

Your calculation is correct: you find the deceleration and then use f= ma.
 

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