In the figure shown, a spring of spring constant `K` is fixed at on end and the other end is attached to the mass 'm' . The coefficient of friction between block and the inclined plane is `mu`. The block is released when the spring is in tis natural length. Assuming that the `theta gt mu`, the maximum speed of the block during the motion is.

Options

A. `(cos theta + mu sin theta) g sqrt((m)/(k))`
B. `(cos theta - mu sin theta) g sqrt((m)/(k))`
C. `(sin theta + mu cos theta) g sqrt((m)/(k))`
D. `(sin theta - mu cos theta) g sqrt((m)/(k))`

Explanation

The speed is maximum when acceleration is least
Let displacment of block is `x_(0)` when the speed of block maximum.

At displacement , applying Newton's law to the block along the incline
`mg sin theta = mu mg cos theta + kx_(0)` (1)
Applying work energy theorem to block along and final position is
`k_(f) = k_(i) + mg x_(0) sin theta - (1)/(2) kx_(0)^(2) - mu mg x_(0) cos theta` (2)
Solving (1) and (2) we get,
`V_(max) = (sin theta - mu cos theta) g sqrt((m)/(k))`