A student proposes a design for an automobile crash barrier in which a 1650-kg s

Question

A student proposes a design for an automobile crash barrier in which a 1650-kg sport utility vehicle moving at 19.0 m/s crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00g.

A) Find the required spring constant k. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground.

B)Find the distance the spring will compress in slowing the vehicle to a stop.

Explanation / Answer

m = mass of automobile = 1650 kg

Vi = initial speed = 19 m/s

Vf = final speed = 0 m/s

a = acceleration = -5 g = -5 x 9.8 = -49 m/s2

d = distance through which spring is compressed

Using the equation :

Vf2 = Vi2 + 2 a d

02 = 192 + 2 (-49) d

d = 3.68 m

using the conservation of energy

Kinetic energy of car = spring potential energy

(0.5) m Vi2 = (0.5) k d2

1650 (19)2 = k (3.68)2

k = 43984.1 N/m

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