AJ22
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- Homework Statement
- Given the change in kinetic energy and distance, find the constant force needed in the crumple zone to stop the vehicle.
- Relevant Equations
- Work Energy Theorem: W = ΔEk
W = FcosxΔd
The vehicle comes to a stop after the collision so my kinetic energy is equal to (2000kg)(22m/s)2 / 2.
(Ekfinal -Ekinital)
I used the equation FcosxΔd = ΔEk. Knowing that the kinetic energy is -484000J and the length of the crumple zone is 0.70m, I can substitute those values into the equation.
Fcosx(0.70) = -484000.
My question is what would be the angle in the equation of W = FcosxΔd. During a collision the force from a wall acts in the opposite direction from the force of the crumple zone? So the angle is cos180 degrees = -1. Is this right?
So the constant force would then be F = -484000/-0.70 = 691Kn?
That means this number is also the crumple zone rigidity.
(Ekfinal -Ekinital)
I used the equation FcosxΔd = ΔEk. Knowing that the kinetic energy is -484000J and the length of the crumple zone is 0.70m, I can substitute those values into the equation.
Fcosx(0.70) = -484000.
My question is what would be the angle in the equation of W = FcosxΔd. During a collision the force from a wall acts in the opposite direction from the force of the crumple zone? So the angle is cos180 degrees = -1. Is this right?
So the constant force would then be F = -484000/-0.70 = 691Kn?
That means this number is also the crumple zone rigidity.
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