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(B) Let the velocity of block $A$ be $v_A = 10 \, m/s$ downwards. The movable pulley connected to block $A$ moves downwards with the same velocity $v_

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Let the velocity of block $A$ be $v_A = 10 \, m/s$ downwards. The movable pulley connected to block $A$ moves downwards with the same velocity $v_A = 10 \, m/s$.Let the velocity of block $B$ be $v_B$ along the horizontal surface.The velocity of the string segment connected to block $B$ is $v_B$. The component of this velocity along the string towards the pulley is $v_B \cos 37^{\circ}$.For the movable pulley,the velocity of the center is $v_A = 10 \, m/s$. The velocity of the string on the left side is $0$ (fixed end),and the velocity of the string on the right side is $v_B \cos 37^{\circ}$.Using the relation for the velocity of the center of a pulley: $v_{pulley} = \frac{v_1 + v_2}{2}$.Here,$10 = \frac{0 + v_B \cos 37^{\circ}}{2}$.$v_B \cos 37^{\circ} = 20$.Since $\cos 37^{\circ} = \frac{4}{5}$,we have $v_B 	imes \frac{4}{5} = 20$.$v_B = \frac{20 	imes 5}{4} = 25 \, m/s$.

Find the velocity of block $B$ at the instant shown in the figure (in $m/s$).

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