Question

# A block A of mass 7 kg is placed on a frictionless table A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (given $g=10m{s}^{-2}$)

A

$100m{s}^{-2}$

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B

$3m{s}^{-2}$

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C

$10m{s}^{-2}$

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D

$30m{s}^{-2}$

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Solution

## The correct option is B $3m{s}^{-2}$Step1: Analyzing the given diagramLet a be the system's acceleration, and T be the tension force acting on the sting.Then block B will begin to move in a downward direction, while block A will begin to move in a forward direction.As illustrated in the diagram. Since tension is the force acting on the string, the forces on block B are ${m}_{B}g,{m}_{B}a$, and the forces on block A are ${m}_{A}a$.Step2: Given dataMass of block A, ${m}_{A}=7kg$Mass of block B, ${m}_{B}=3kg$$g=10m{s}^{-2}$Step3: Formula used$T=ma\left(whereT=tension,m=mass,a=acceleration\right)$Step4: Calculating the tensionFor block A, the tension force is the only force that allows block A to move forwardSo $T={m}_{A}a.........\left(i\right)$ is the result of balancing both forces. For block B, we can see that the weight and motion forces ${m}_{B}g,{m}_{B}a$ are acting in a downward direction, which is balanced by tension T acting on the string, which can be written as ${m}_{B}a={m}_{B}g-T.......\left(ii\right)$Step5: Calculating accelerationFrom equation (i) put the value of $T={m}_{A}a$ we get,$\left({m}_{B}+{m}_{A}\right)a={m}_{B}g\phantom{\rule{0ex}{0ex}}\left(3+7\right)a=3×10\phantom{\rule{0ex}{0ex}}10a=30\phantom{\rule{0ex}{0ex}}a=3m{s}^{-2}$Hence, option B is the correct answer.

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