A5.00×105-kg subway train is brought to a stop from a speed of 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

This can be calculated with the law of conservation of energy. The kinetic energy Ek of the train is completely transformed to potential energy of the spring Ep which stops the train:

Ek=Ep

(1/2)*m*v²=(1/2)*k*x², where m is mass of the train, v is the velocity of the train, k is the force constant of the spring and x is the path the train went while being stopped and also the amount of length the spring compressed. 
We solve for k:

(1/2)*m*v²=(1/2)*k*x², 1/2 cancel out:

m*v²=k*x², we divide both sides by x² and get k:

k=(m*v²)/x²={(5*10^5)*(0.5²)}/(0.4²)=781250 N/m

So the force constant of the spring is: k=781250 N/m.