Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. cars are designed with a "crumple zone" in the front of the car. in the event of an impact, the passenger compartment decelerates with the car. by contrast, an unrestrained occupant keeps moving forward with no loss of speed (newton's first law! ) until hitting the dashboard or windshield, as we saw in figure 4.2. these are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about 5 mm.
a. a 60 kg person is in a head-on collision. the car's speed at impact is 15 m/s.
estimate the net force on the person if he or she is wearing a seat belt and if the air
bag deploys.
b. estimate the net force that ultimately stops the person if he or she is not restrained
by a seat belt or air bag.
c. how does these two forces compare to the person's weight

a)   = 1.35 10⁴ N , b)   F = 1.35 10⁶ N, c)  F /    = 100

Explanation:

a) Let's start by calculating the aceleration it takes for the person with a seat belt and air bag to stop, the approximate distance between the floor and the steering wheel is about 50 cm, let's use kinematics to calculate the acceleration

       v² = v₀² - 2 a x

       0 = v₀² - 2 a x

       a = v₀² / 2 x

       a = 15² / (2 0.50)

       a = 225 m / s²

We calculate the force with Newton's second law

       F = m a

       F = 60 225  

        = 1.35 10⁴ N

b) we perform the same calculation for a person without a belt

      a = v₀² / 2x

      a = 15² / (2 0.005)

      a = 22500 m / s²

      F = m a

      F = 60 22500

     F = 1.35 10⁶ N

c) let's calculate the relationship between these two forces

      F / = 1.35 10⁶ / 1.35 10⁴

      F /    = 10²

we need to know the time interval during which the force acts. impulse of force = momentum change

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards