Force and Newton's laws (airbag)

[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?

 

[HallsofIvy]

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.

 

[M98Ranger]

Yes, your approach is correct. The force acting on the passenger's upper torso can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the acceleration is negative because the person is decelerating due to the impact with the airbag. This negative acceleration results in a negative force, but since we are looking for the magnitude of the force, we can take the absolute value to get the final answer. Keep in mind that the negative sign just indicates the direction of the force, which in this case is opposite to the direction of motion. Great job on using the appropriate equations and units!