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A car traveling at 53Km/hhits a bridge abutment. A passenger in the car moves forward a distance of65cm(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鈥檚 upper torso, which has a mass of41kg?

Short Answer

Expert verified

The magnitude of force acts on a passenger is-6.8103N.

Step by step solution

01

The given data

  • Speed of the car, vi=53km/hor14.72m/s.
  • Distance traveled by the car,x=65cmor0.65m.
  • Mass of passengers, m=41kg.
02

Understanding the concept of Newton’s 2nd law

As we know the initial and final speed of the passenger and the distance traveled by him, using kinematic equations, we can find the acceleration (Deceleration) of the person.

Using this acceleration and Newton鈥檚 2nd law we can find the force acting on it.

Formula:

The third kinematic equation of motion,vi2=vf2+2ax (i)

The force according to Newton鈥檚 second law,F=ma (ii)

03

Calculation of magnitude of force

Using the given values and equation (i), we get the acceleration of the car as:

(14.7m/s)2=0+2a0.65m216.09=1.3aa=-216.11.3=-166.22m/s2

Now, using equation (ii) and the given values, we get the magnitude of force as follows:

F=41kg(-166.22m/s2)=-6815.52N=-6.8103N

Negative sign shows that the force is acting in the opposite direction of the motion and the magnitude value of force is6.8103N .

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