Question

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?

Answer 1

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