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23 January, 22:06

A car traveling at 43 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 54 cm (with 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 49 kg?

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  1. 23 January, 23:35
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    F = - 6472.9 N

    F = - 6.47 kN

    Explanation:

    First of all you have to convert the data to SI units

    so for the velocity you have:

    Vi = 43km/h * (1000m/1km) * (1h/3600s) - --> using conversion factors

    Vi = 11.9444 m/s

    dX : distance the passanger moves

    dX = 54cm * (1m/100cm) - -> using conversion factors

    dX = 0.54 m

    Now to calculate the force we are going to use the sum of focers equals to mass for acceleration:

    Sum F = m*a

    We have to find a so we are going to use the velocity's formula as follows to solve a:

    Vf ^2 = Vi^2 + 2*a*dX

    Vf=0 - -> the passenger does not move after the airbag inflates.

    a = - (Vi^2) / (2*dX)

    you solve de acceleration with the data you hae and you will find

    a = - 132.1 m / s^2

    Now you can solve the Sum F equation

    Sum F = 49 Kg * (-132.1 m/s^2)

    F = - 6472.9 N

    F = - 6.47 kN
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