A car has a crumple zone that is 0.80 m (80 cm) long. In this car, the distance from the dummy to the steering wheel is 0.50 m. The car has a mass of 1,600 kg and the dummy has a mass of 75 kg. At the time of the crash, the car has a speed of 18 m/s. Based on the work-energy theorem, what is the smallest possible force that the dummy could experience during the crash

F = 518.4 KN

Explanation:

From work energy theorem;

Energy = workdone

In this case, Energy is kinetic energy and has a formula ½mv²

Thus;

Fd = ½mv²

We are given;

Mass of car; m = 1600 kg

Speed of car after crash; v = 18 m/s

Distance from dummy to steering wheel; d = 0.5 m

Thus;

½ × 1600 × 18² = F × 0.5

This gives;

800 × 324 = 0.5F

F = 800 × 324/0.5

F = 518400 N

F = 518.4 KN

A -9.35

Explanation: The work-energy theorem states the work (Fd) is equal to the change in kinetic energy: Fd = ΔKE. Before the crash, the kinetic energy of the dummy is equal to one-half mass times velocity squared, or 0.5 · 75 kg · (18 m/s)2, which is equal to 12,150 J. The kinetic energy of the dummy after the crash is 0 J, so the change in kinetic energy is –12,150 J. If the crumple zone collapses completely and the dummy just misses hitting the steering wheel, then the distance d is 0.80 + 0.50 = 1.30 m. So we have:

F · d = ΔKE

F = ΔKE/d

F = –12,150 J/1.30 m = 9,346 N, or 9.35 kN

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more