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(C) Let $L$ be the length of the string segment from the pulley to block $B$. Let $x$ be the horizontal distance of block $B$ from the point directly

Vedclass · Accepted Answer

$\frac{u}{\cos 	heta}$

Vedclass · Answer

(C) Let $L$ be the length of the string segment from the pulley to block $B$. Let $x$ be the horizontal distance of block $B$ from the point directly below the pulley.From the geometry of the figure,the length of the string $L$ is related to the horizontal distance $x$ by $x = L \cos 	heta$.Since the string is inextensible,the rate of change of the total length of the string is constant. The velocity of block $A$ is $u$,which means the string is being pulled at a rate of $u$.Let $v$ be the velocity of block $B$. The component of the velocity of block $B$ along the string must be equal to the velocity of block $A$.Therefore,$v \cos 	heta = u$.Solving for $v$,we get $v = \frac{u}{\cos 	heta}$.

Two equal masses $A$ and $B$ are arranged as shown in the figure. The pulley and string are ideal and there is no friction. Block $A$ has a speed $u$ in the downward direction. The speed of the block $B$ is:

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