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

Explanation

Given, mass of the system(m)=50,000kg
Speed of the system (v)=36km/h
`=(36xx1000)/(60xx60)=10m//s`
Compression of the spring(x)=1.0m
KE of the system=`(1)/(2)mv^(2)`
`(1)/(2)xx50000xx(10)^(2)`
`=25000xx100J=2.5xx10^(6)`
Since, 90% of KE of the system is lost due to friction, therefore, energy transferred to shock absorber, is given by
`DeltaE=(1)/(2)kx^(2)=10%` of total KE of the system
`=(10)/(100)xx2.5xx10^(6)J or k=(2xx2.5xx10^(6))/(10xx(1)^(2))`
`=5.0xx10^(5)N//m`