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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?

Respuesta :

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

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