HRW5 10-3E The National Transportation Safety Board is testing the crashworthiness of a new car. The 1.36 103 kg vehicle, moving at 15.6 m/s, is allowed to collide with a bridge abutment, being brought to rest in a time of 0.465 s. What force, assumed constant, acted on the car during impact?
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2007-02-08 11:27:40 · 4 answers · asked by ckielblock18 1 in Science & Mathematics ➔ Physics
impulse is changing momentum, when the car is moving, it has momentum, in order to bring the car to stop, impulse has to take away all of the momemtum
Ft = mv
F(.465) = (1.36x10^3kg)(0 - 15.6)
F(.465) = -21216
F = -45,625.8N
the negative means the force is opposite of the motion.
2007-02-08 11:46:52 · answer #1 · answered by 7 · 0⤊ 0⤋
This problem can be solved in many ways, but they're all based on the second Newton's law:
F = ma = m x Δv/Δt = m x v / t = 1360 kg x 15.6 m/s / 0.465 s =
= 45.6 kN
2007-02-08 11:35:23 · answer #2 · answered by Dorian36 4 · 0⤊ 0⤋
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2016-12-17 05:36:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Sorry, thats a crazy subject. I'm glad I never took it :)!! Good luck though !
2007-02-08 11:35:20 · answer #4 · answered by Anonymous · 0⤊ 0⤋