In the figure shown, a spring of spring constant k is fixed at one end and the other end is attached to the mass 'm'. The coefficient of friction between block and the inclined plane is 'μ'. The block is released when the spring is in its natural length. Assuming that tan θ>μ, find the maximum speed of the block during the motion.
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a
(sin θ+μ cos θ) gmk
b
(sin θ-μ cos θ) gmk
c
(cos θ−μ sin θ) gmk
d
(cos θ+μ sin θ) gmk
answer is B.
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Detailed Solution
The speed is maximum when acceleration is zero.Let displacement of block is x0 when the speed of block is maximum.At equilibrium, applying Newton's law to the block along the incline mg sin θ=μ mg cos θ+kx0(mg sin θ−μ mg cos θ)=kx0 …(i)Applying work energy theorem to block between initial and final position is(KE)f=(KE)i+mg x0 sin θ−12kx02−μ mg x0 cos θ12mvmax2=0+mg x0 sin θ−12kx02−μ mg x0 cos θ … (ii) 12mvmax2=(mg sin θ−μ mg cos θ)x0−12kx02Solving (i) and (ii) we get,vmax=(sin θ−μ cos θ) gmk