In the position 1 (figure) the spring of constant K is undeformed. Find the work done by the force $\vec{F}$ (which is always directed along the tangent to the smooth hemi spherical surface) on the small block of mass m to shift it from the position 1 to position 2 slowly.

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mg R $\sin θ_{0} + \frac{K R^{2} θ_{0}^{2}}{2}$

mg R $\sin θ_{0} + \frac{K R^{2} θ_{0}^{2}}{4}$

mg R $\sin θ_{0}$

Zero

answer is A.

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Detailed Solution

Let us locate the block at an arbitrary angular position $θ < θ_{0}$
$d W = \vec{F} . \vec{d r} = F d s = F (R d θ) ... (1)$
Now from the condition of equilibrium of the block, we have
$F = m g \cos θ + K R θ ..... (2)$
From eq. (1) & (2)
dW= $m g R \cos θ d θ + K R^{2} θ d θ$
Hence the sought work