To simulate car accidents , auto manufactures study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed `18.0` km/h on a smooth road and colliding with a horizontally mounted spring of spring constant `6.25 xx10^(3) Nm^(-1)` .What is the maximum compression of the spring ?

Explanation

At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring .
The kinetic energy of the moving car is ,
`K = 1/2 mv^(2)`
`=1/2 xx10^(3)xx(5)^(2)`
` = 1.25 xx10^(4)J`
`v = 18 "kmh"^(-1)`
` = (18 xx1000)/(3600)`
` = 5 ms^(-1)`
From the law of the conservation of mechanical energy potential energy of spring = kinetic energy of moving car `V = 1/2 kx_(m)^(2)`
Where k = spring constant , `x_(m) ` = maximum compression
` 1.25 xx10^(4) =1/2 xx6.25 xx10^(3) xxx_(m)^(2)`
` =(2xx1.25 xx10^(4))/(6.25 xx10^(3)) =x_(m)^(2)`
` :. x_(m) = sqrt((2.5xx10^(4))/(6.25xx10^(3))`
` sqrt(4)`
` :. x_(m) = 2m `
` :. ` Since spring will be compressed by 2 m .