Barrier Deformation and Impact of Car

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SUMMARY

The discussion focuses on calculating the barrier deformation required to bring a car to rest after a collision. The force exerted on the car by the crash barrier is defined by the equation F=-(4.5+140s) kN, where 's' represents the distance in meters from the initial contact. A car with a mass of 2000 kg traveling at 100 km/h requires a barrier deformation of approximately 3.29 meters to stop. The work-energy theorem is applied to equate the work done by the barrier to the initial kinetic energy of the car.

PREREQUISITES
  • Understanding of the work-energy theorem
  • Familiarity with impulse-momentum principles
  • Knowledge of basic calculus for integration
  • Ability to convert units (e.g., km/h to m/s)
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn about impulse-momentum relationships in collisions
  • Practice integration techniques for calculating work done
  • Explore examples of barrier deformation calculations in crash simulations
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and safety analysts involved in vehicle crash testing and barrier design.

depre87
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Homework Statement


Force exerted on a car by a crash barrier as the barrier crushes is F=-(4.5+140s) kN where s is the distance in metres from the initial contact. If a car of mass 2000 kg is traveling at 100 km/h when it hits the barrier the barrier deformation required to bring the car to rest will be most nearly: 0.35,1.04,1.78,2.81,3.29 metres

Homework Equations


Not entirely sure which to use here? Impulse => Fdt=mv1-mv0 or conservation of energy:
0.5mv^2+(F)ds = 0

The Attempt at a Solution


0.5mv^2+(F)ds = 0
0.5(2000)(27.78)^2+[-4.5s-70s^2] = 0, solve for s and i believe i got the incorrect answers (it is supposed to be 3.29m)

any clues guys am i completely off in my method? thanks!
 
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Your method looks fine. If you use the work-energy theorem, you get W = ΔK. To bring the car to rest, the final K must be 0, so W = 0 - K0 = -K0 where K0 is the initial kinetic energy. Let S (capital) be the displacement of the barrier that is enough to bring the car to a rest. Then

-K_0 = W = \int_0^S F(s)\,ds = -\int_0^S (4.5 + 140s)\,ds

K_0 = 4.5S + 70S^2

After this, it's just a matter of plugging in the given values for S and seeing which one gets you closest to K0. If you got it wrong, you must have an arithmetic error somewhere. What did you get for K0?
 

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