Question

A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance of 56.0 cm while being brought to rest by an inflated air bag. Assuming that the force that stops the passenger is constant, what is the magnitude F of this force?

Answer 1

`F=43570.9N`

Step-by-step explanation:

We can calculate the acceleration experimented by the passenger using the formula
`v_f^2=v_i^2+2ad`, taking the initial direction of movement as the positive direction and considering it comes to a rest:

`a=(v_f^2-v_i^2)/(2d)=(-v_i^2)/(2d)`

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

`F=ma=(-mv_i^2)/(2d)`

Which for our values is:

`F=(-(65kg)(27.4m/s)^2)/(2(0.56m))=43570.9N`