In the figure shown the pulley is smooth . The spring and the string are light. The block `'B'` slides down from the top along the fixed rought wedge of inclination `theta`. Assuming that the block reaches the end of the wedge. Find the speed of the block at the end. Take the coefficient of friction between the block and the wedge to be `mu` and the spring was relaxed when the block was relased from the top of the wedge.

Answer

`V=sqrt((2)/(m)[mgh-(1)/(2K)((h)/(sin theta))^(2)-mu mgh cot theta])`

Explanation

By energy conservation `:`
`mgh=(1)/(2)mv^(2)+(1)/(2)K((h)/(sin theta))^(2)+ mu mg cos theta. (h)/(sin theta )rArr V`
`sqrt((2)/(m)[mgh-(1)/(2)K((h)/(sin theta))^(2)-mu mgh cot theta])Ans.`