At a given instant,block A is moving with a v | vedclass

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(A) Let the total length of the string be $L$. Based on the pulley arrangement,the length of the string can be expressed as $L = x_A + 2x_P + x_B$,whe

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$15\,m/s\downarrow$

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(A) Let the total length of the string be $L$. Based on the pulley arrangement,the length of the string can be expressed as $L = x_A + 2x_P + x_B$,where $x_A$ is the position of block $A$,$x_P$ is the position of the movable pulley,and $x_B$ is the position of block $B$.Since block $A$ is attached to the movable pulley,$x_A = x_P$. Thus,$L = 3x_A + x_B$.Differentiating with respect to time $t$,we get $0 = 3v_A + v_B$.Given that block $A$ is moving upwards with $v_A = -5\,m/s$ (taking upwards as negative and downwards as positive).Substituting the value: $0 = 3(-5) + v_B$.$v_B = 15\,m/s$.Since the result is positive,block $B$ is moving downwards. Therefore,the velocity of block $B$ is $15\,m/s\downarrow$.

At a given instant,block $A$ is moving with a velocity of $5\,m/s$ upwards. What is the velocity of block $B$ at that time?

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