what must be the effective spring constant of the bumper material, assuming the car has a mass of 1110 kg and is tested by ramming into a solid wall?

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 2.3 cm, what must be the effective spring constant of the bumper material, assuming the car has a mass of 1110 kg and is tested by ramming into a solid wall?

Answer 1

The force must this case must be the force the brings the car to a stop, it must be also equal to the critical elastic force.

F = ma = kx

We are given a mass (1110kg) and a compression (2.3cm), but we need to know the acceleration to find k.

The other information in the question leads us to use one of the equations of motion.

Braking implies that the velocity goes from v_o = 8 km/h = 8000m/3600s to v = 0 over this critical distance.

0 = (8000/3600)² + 2a (0.023) ⇒ a = -107.35 m/s²

Now subsituting in the force relation ma = -kx

(1110)(107.35) = k(0.23) ⇒ k =(1110)(107.35)/0.23 = 5180000N/m

If the number looks big remember you need a good amount of stiffness (and low speed given) for the car to not get smashed.