In the figure shown,the block B moves down wi | vedclass

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(B) Let the velocity of block $B$ be $v_B = 10 \ m/s$. The block $B$ is attached to a movable pulley. The velocity of the string segment connected to

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$25$

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(B) Let the velocity of block $B$ be $v_B = 10 \ m/s$. The block $B$ is attached to a movable pulley. The velocity of the string segment connected to block $A$ is $v_s = 2 	imes v_B = 2 	imes 10 = 20 \ m/s$.Let $x$ be the horizontal distance of block $A$ from the pulley and $l$ be the length of the string segment from the pulley to block $A$. From the geometry,$l = x \sec(37^{\circ})$.Differentiating with respect to time $t$,we get $\frac{dl}{dt} = \frac{dx}{dt} \sec(37^{\circ})$.Here,$\frac{dl}{dt} = v_s = 20 \ m/s$ and $\frac{dx}{dt} = v_A$ (velocity of block $A$).Thus,$v_A = v_s \cos(37^{\circ}) = 20 	imes \frac{4}{5} = 16 \ m/s$.Wait,re-evaluating the constraint: The component of the velocity of block $A$ along the string must equal the velocity of the string.$v_A \cos(37^{\circ}) = v_s$$v_A 	imes (4/5) = 20$$v_A = 20 	imes (5/4) = 25 \ m/s$.

In the figure shown,the block $B$ moves down with a velocity of $10 \ m/s$. The velocity of $A$ in the position shown is ......... $m/s$.

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