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

Answer:

$F=43570.9N$

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:

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}$

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

$F=ma=\frac{-mv_i^2}{2d}$

Which for our values is:

$F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N$