Today\'s cars have elastic bumpers that are designed to compress and rebound wit

Question

Today's cars have elastic bumpers that are designed to compress and rebound without any physical damage at speeds below about 5 mi/h (8 km/h). The material of the bumpers behaves essentially as an ideal spring up to that point but permanently deforms beyond that. If the compression corresponding to the elastic limit for a particular bumper is 1.5 cm, what must be the effective spring constant of the bumper material, assuming the car has a mass of 1150 kg and is tested by ramming into a solid wall?
N/m

Explanation / Answer

The compression in spring in the elastic regime obey's Hooke's law which states that the force needed to extend or compress a spring by some distance is proportional to that distance, i.e. F = kx.

Correspondingly, the elastic potential energy stored in the spring is -

U = (1/2)kx2

Now, initial kinetic energy of bumper is:

E = (1/2)mv2

This kintetic energy is completely transferred into spring potential energy at maximum compression.

i.e. U = E

or, (1/2)mv2 = (1/2)kx2

or, k = mv2/x2

Given, m = 1150 kg, v = 8 km/h = 2.22 m/s, x = 1.5 cm = 0.015 m

Therefore,

k = mv2/x2 = 1150 x (2.22)2/(0.015)2 = 2.524 x 107 N/m

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