As a civil engineer for your city, you have been assigned to evaluate the purchase of spring-loaded guard rails to prevent cars from leaving the road. In response to a request for proposals*, one company states their guard rails are perfect for the job. Each section of their guard rails consists of two springs, each having a force constant 3.13400 105 N/m with a maximimum distance of compression of 0.614 m. (According to the manufacturer, beyond this compression the spring loses most of its ability to absorb an impact elastically.) The largest vehicle the guardrails are expected to stop are trucks of mass 4550.000 kg. What is the maximum speed at which these guard rails alone can be expected to bring such vehicles to a halt within the stated maximum compression distance? (Assume the vehicles can strike the guard rail head on and that the springs are perfectly elastic.)
m/s
Given your result which section of road often features such a speed
School Zone
Large Road
Highway
Guard rails are pointless if the acceleration they create seriously injures passengers. One important safety factor is the acceleration experienced by passengers during a collision. Calculate the maximum acceleration of the vehicle during the time in which it is in contact with the guard rail.
_ m/s^2
If the highway department considers 20 g\'s the maximum safe acceleration, is this guard rail safe in regards to acceleration?

We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression

Initial

Em₀ = K = ½ m v²

Final

[tex]Em_{f}[/tex] = 2 Ke = ½ k x²

The two is placed because each barred has two springs and each does not exert the same force

Emo = [tex]Em_{f}[/tex]

½ m v² = 2 ½ k x²

v = √(2k/m) x

v = √ (2 3,134 10⁵/4550) 0.614

v = 7,207 m / s

Let's take this speed to km / h

v = 5,096 m / s (1km / 1000m) (3600s / 1h)

v = 25.9 km / h

This speed is common in school zones

Let's use kinematics to calculate the average acceleration

vf² = v₀² - 2 a x

0 = v₀² - 2 a x

a = v₀² / 2 x

a = 7,207²/2 0.614

a = 42.3 m / s²

We buy this acceleration with the acceleration of gravity

a / g = 42.3 / 9.8

a / g = 4.3

This acceleration is well below the maximum allowed