An engine is attached to a wagon through a shock absorber of length `1.5 m`. The system with a total mass of 50,000 kg is moving with a speed of `36 ^kmh(-1)` when the brakes are applied to bring it to rest, the spring of the shock absorber gets compressed by 1.0 m. If `90 % ` of energy of the wagon is lost due to friction, calculate the spring constant.

Explanation

m = 50,00 kg
v = 36 km/h
` = (36xx1000)/(60xx60)`
= 10 m/s
x = 1.0 m
` K = 1/2 mv^(2)`
` = 1/2 xx 50000 xx(10)^(2)`
` = 25000 xx100 J `
`= 2.5 xx10^(6)` J
Since , 90 % of KE of the system is lost due to friction , therefore 10 % energy transferred to shock ,absorber , is given by
`Delta E = 1/2 kx^(2)`
`DeltaE = 10 % of K `
` :. 1/2 kx^(2) = 10/10 xx2.5 xx10^(6) J `
` :. k = 5.0 xx10^(5) ` N/m