A train wagon is attached to the engine through a shock absorber 1.5 m long the system having a total mass of 5 x 104 kg is going at a speed of 36 km/h. Suddenly the brakes are applied to bring them to rest. In the process of coming to rest, the spring of the shock absorber gets compressed by 1 metre. If only 90% energy of the wagon is lost due to friction. What is spring constant of spring?

Options

A. `5xx10^(4)` N/m
B. `10^(5)` N/m
C. `10^(4)` N/m
D. `5xx10^(5)` N/m

Answer

d

Explanation

Here speed of wagon = `36 km h^(-1)`
`v=36xx(5)/(18)=10 ms^(-1)`
`:.` K.E. of system =`1//2 Mv^2`
`E_k=1/2xx5xx10^4xx10^2`
=`2.5xx10^6 J`
Now 90% of this is lost due to friction.10% of this is used for compressing the spring i.e.
`1/2kx^2=(2.5xx10^6xx10)/(100)`
=`2.5xx10^5 J`
or `1/2xxkx(1)^2=2.5xx10^5`
`:. K=5xx10^5 N//m`