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Q. In the following arrangement the system is initially at rest. The $5 \,kg$ block is now released Assuming the pulleys and string to be massless and smooth, the acceleration of block ā€˜C’ will be
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Laws of Motion

Solution:

Block $B$ will not move
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$5g-T=5a \ldots$ (1)
$2T-8g=8 \frac{a}{2} \ldots$ (2)
$10g-2T=10a$ from (1)
$2g=14a$
$a=\frac{g}{7}$
$\therefore \frac{a}{2}=\frac{g}{14}=\frac{10}{14}$
$=\frac{5}{7} m/s^{2}$