A 1200 kg car rolling on a horizontal surface has speed v = 50 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

When the car is moving, it has Kinetic Energy. As soon as the car entercount with the spring, its Kinetic Energy will transfer to Elastic Potential Energy.

KE = EPE

1/2mv^2 = 1/2kx^2

change the speed into m/s

50 km/hr = 13.9 m/s

1/2 (1200) (13.9) ^2 = 1/2k (2.2) ^2

now you do the math to solve for k.

k = 115926/2.42

K = 47903.30