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. (a) `(sin theta+mu cos theta)gsqrt(m/k)`
B. (b) `(sin theta-mu cos theta)g sqrt(m/k)`
C. (c) `(cos theta-mu sin theta)g sqrt(m/k)`
D. (d) `(cos theta+musintheta)gsqrt(m/k)`

Answer

B

Explanation

The speed is maximum when acceleration is zero.
Let displacement of block is `x_0` when the speed of block is maximum.
At equilibrium, applying Newton's law to the block along the incline `mg sin theta=mu mg cos theta+kx_0`
`(mg sin theta-mumg cos theta)=kx_0` ...(1)
Applying work energy theorem to block between initial and final position is
`(KE)_f=(KE)_i+mg x_0 sin theta-1/2kx_0^2-mu mg x_0 cos theta`
`1/2mv_(max)^2=0+mgx_0sin theta-1/2kx_0^2-mumgx_0cos theta`
...(2)
`1/2mv_(max)^2=(mg sin theta-mu mg cos theta)x_0-1/2kx_0^2`
Solving (1) and (2) we get,
`v_(max)=(sin theta-mu cos theta)gsqrt(m/k)`