A car’s bumper is designed to withstand a 6.84 km/h (1.9-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.285 m while bringing a 830 kg car to rest from an initial speed of 1.9 m/s.
Answer:
F = 5.256 x [tex]10^{3} N[/tex]
Explanation:
From the work energy theorem we know that:
The net work done on a particle equals the change in the particles kinetic energy:
W = F.d, ΔK =[tex]\frac{1}{2} mv^{2}_{f} – \frac{1}{2} mv^{2}_{i} , F.d = \frac{1}{2}mv^{2}_{f} -\frac{1}{2} mv^{2}_{i}[/tex]
where:
W = work done by the force
F = Force
d = Distance travelled
m = Mass of the car
vf, vi = final and initial velocity of the car
kf, ki = final and initial kinetic energy of the car
Given the parameters;
m = 830kg
vi = 1.9 m/s
vf = 0 km/h
d = 0.285 m
Inserting the information we have:
F.d = [tex]\frac{1}{2} mv^{2}_{f} – \frac{1}{2} mv^{2}_{i}[/tex]
F = [tex]\frac{\frac{1}{2} mv^{2}_{f} – \frac{1}{2} mv^{2}_{i} }{d}[/tex]
F = [tex]\frac{ 0 – \frac{1}{2} X830 X 1.9^{2} }{0.285}[/tex]
F = 5.256 x [tex]10^{3} N[/tex]