Answer:
The net force is 1350 kN
Solution:
As per the question:
Mass of man, m = 60 kg
Initial speed of the car, v = 15 m/s
Final speed of the car, v' = 0 m/s
Distance covered by the person before coming to rest, d = 5 mm = [tex]5\times 10^{- 3}\ m[/tex]
By using the third eqn of motion:
[tex]v'^{2} = v^{2} + 2ad[/tex]
[tex]0 = 15^{2} + 2\times a\times 5\times 10^{- 3}[/tex]
[tex]a = - 22500\ m/s^{2}[/tex]
Thus the force on the person can be given by:
F = ma = [tex]60\times 22500 = 1350\ kN[/tex]