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Byju's Answer

Standard XII

Physics

Problem Solving

Three blocks ...

Question

Three blocks are connected by strings as shown in figure and pulled by a force $F$ $=$ $60 N$ . If $m_{A} =$ $10 k g$ , $m_{B} =$ $20 k g$ , and $m_{C} =$ $30 k g$ , then:

acceleration of the system is

2

m / s^{2}

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T_{1} = 10 N

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T_{2} = 30 N

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T_{1} = 20 N

and

T_{2} = 40 N

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Solution

The correct options are
A $T_{1} = 10 N$
C $T_{2} = 30 N$
Total force F is dragging the combined weight of A, B and C.
Therefore, $60 = (10 + 20 + 30) a$
$\Rightarrow a = 1 m / s^{2}$
Tension $T_{2}$ is dragging the weights of A and B.
Therefore, $T_{2} = 30 N$
Similarly, Tension $T_{1}$ is dragging the weights of A only.
Therefore, $T_{1} = 10 N$

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Three blocks of mass $M_{1}$ , $M_{2}$ & $M_{3}$ are connected by massless strings as shown on a frictionless table. Where $T_{1}$ is the tension in the string between block $M_{1}$ and $M_{2}$ , $T_{2}$ is the tension in the string between block $M_{2}$ and $M_{3}$ . They're pulled with a force $F = 40 N$ . If $M_{1}$ =10 kg, $M_{2}$ = 6 kg and $M_{3}$ = 4 kg, then tension $T_{2}$ will be:

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Three blocks are connected by strings as shown in figure and pulled by a force F = 60N. If mA=10 kg, mB=20 kg, and mC=30 kg, then:

The correct options are A T1=10N C T2=30NTotal force F is dragging the combined weight of A, B and C.Therefore, 60=(10+20+30)a⇒a=1 m/s2Tension T2 is dragging the weights of A and B. Therefore, T2=30NSimilarly, Tension T1 is dragging the weights of A only. Therefore, T1=10N

Three blocks are connected by strings as shown in figure and pulled by a force $F$ $=$ $60 N$ . If $m_{A} =$ $10 k g$ , $m_{B} =$ $20 k g$ , and $m_{C} =$ $30 k g$ , then: