Today\'s cars have elastic bumpers that are designed to compress and rebound wit
ID: 585838 • Letter: T
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