A 5.00×105-kg subway train is brought to a stop from a speed of

Question

0.500 m/s in 0.400 m by a large spring bumper at the end of its track.
What is the force constant k of the spring?

Answer 1

KE=PE
mv²/ 2 =kx²/2
k=m(v/x) ²

Answer 2

mvi^2=kxf
k=mvi^2 / xf2
k=(5.00*105)(0.5)^2 / (0.4)
k=328 N/m

Answer 3

oops, forgot to square my x