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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=10ms-2)


A

100ms-2

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B

3ms-2

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C

10ms-2

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D

30ms-2

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Solution

The correct option is B

3ms-2


Step1: Analyzing the given diagram

  1. Let a be the system's acceleration, and T be the tension force acting on the sting.
  2. Then block B will begin to move in a downward direction, while block A will begin to move in a forward direction.
  3. As illustrated in the diagram. Since tension is the force acting on the string, the forces on block B are mBg,mBa, and the forces on block A are mAa.

Step2: Given data

Mass of block A, mA=7kg

Mass of block B, mB=3kg

g=10ms-2

Step3: Formula used

T=ma(whereT=tension,m=mass,a=acceleration)

Step4: Calculating the tension

  1. For block A, the tension force is the only force that allows block A to move forward
  2. So T=mAa.........(i) is the result of balancing both forces.
  3. For block B, we can see that the weight and motion forces mBg,mBa are acting in a downward direction, which is balanced by tension T acting on the string, which can be written as mBa=mBg-T.......(ii)

Step5: Calculating acceleration

From equation (i) put the value of T=mAa we get,

(mB+mA)a=mBg(3+7)a=3×1010a=30a=3ms-2

Hence, option B is the correct answer.


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