Saved Bookmarks
| 1. |
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? |
|
Answer» `5xx10^(4)` N/m `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` |
|