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A block of mass 1 kg is attached at lower-end of an elastic string (natural length $l_{0} = 1 m$ )
Upper-end of string is slowly pulled till the block just leaves the contact. If at the time of leaving point A on the string (which was originally at a distance 0.8 $l_{0}$ from the mass) shifted by 0.2 $l_{0}$ from its initial position (as shown) then total work done by force in this process is $\frac{x}{y}$ joule (Assume elongation in the string is uniform) then $x + y =____$ .

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answer is 9.

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Detailed Solution

$l_{0} k = k_{1} l_{1} = k_{2} l_{2}$

$l_{1} = 0.2 l_{0}; l_{2} = 0.8 l_{0}$

$T_{1} = k_{1} Δ X_{1 }; T_{2} = k_{2} Δ X_{2}$

$Δ X_{T o t} = Δ X_{1} + Δ X_{2} = \frac{1}{4} m$

$W = \frac{1}{2} k_{1} Δ {X_{1}}^{2} + \frac{1}{2} k_{2} Δ {X_{2}}^{2} = \frac{m g l_{0}}{8} = \frac{10}{8} = \frac{5}{4} = \frac{x}{y} \Rightarrow x + y = 9$

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