In figure, the pulley shown is smooth. The spring and the string are light. Block B slides down from the top along the fixed rough 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 that the spring was relaxed when the block was released from the top of the wedge.

Answer

`sqrt(2/m[mgh-1/2K((h)/(sintheta))^2-mumghcottheta])`

Explanation

Here block will loose the gravtitational potential energy. This energy will be divided in three ways.
a. Some part will be converted in the KE of block
b. Some part will go into spring
c. Remaining part will be used to do work against friction.
By energy conservation:
`mgh=1/2mv^2+1/2K((h)/(sintheta))^2+mumgcos theta((h)/(sin theta))`
`impliesv=sqrt(2/m[mgh-1/2K((h)/(sintheta))^2-mumghcottheta])`