Question

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 1

Answer:

F = 5.256 x $10^{3} N$

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 =$\frac{1}{2} mv^{2}_{f} - \frac{1}{2} mv^{2}_{i} , F.d = \frac{1}{2}mv^{2}_{f} -\frac{1}{2} mv^{2}_{i}$

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 = $\frac{1}{2} mv^{2}_{f} - \frac{1}{2} mv^{2}_{i}$

F = $\frac{\frac{1}{2} mv^{2}_{f} - \frac{1}{2} mv^{2}_{i} }{d}$

F = $\frac{ 0 - \frac{1}{2} X830 X 1.9^{2} }{0.285}$

F = 5.256 x $10^{3} N$