Finding spring constant of bumper

[mohamud3917]

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 1.6 cm, what must be the effective spring constant of the bumper material, assuming the car has a mass of 1070 kg and is tested by ramming into a solid wall

when i tried the k=mg/x
i got k=(1070kg)(9.8)/(0.016m)
so k=655375n/m

but it says its wrong

 

[Doc Al]

You need to consider energy, not force.

 

[mohamud3917]

energy? i am really confused, can you please give me a step by step

thanks

 

[Doc Al]

What you did was find the spring constant needed to stretch 1.6cm if the car where hung from such a spring. But that's not what's going on here. The car has kinetic energy, which must be absorbed by the spring.

How do you find the spring potential energy of a compressed spring?

 

[mohamud3917]

PE_s=1/2kx²

PE_s=1/2(655375n/m)(0.016²m)

PE_s=83.888

so what is next

 

[Doc Al]

PE_s=1/2kx²

This is the right formula for spring PE. Good!

PE_s=1/2(655375n/m)(0.016²m)

This is incorrect. Remember you are solving for the spring constant. (Your calculation from your first post was wrong--otherwise you'd be done.)

You need to set the initial KE equal to the spring PE and solve for the spring constant.

 

[mohamud3917]

ok so how do i find the potential energy of the spring, is it mgh?

 

[Doc Al]

ok so how do i find the potential energy of the spring, is it mgh?

mgh is gravitational PE, which is not relevant here. You gave the correct expression for spring PE in post #5.