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Byju's Answer

Standard IX

Physics

Inertia

Three blocks ...

Question

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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Solution

Given,
Total force applied on the system = 40 N

Total mass of the system= $M_{1} + M_{2} + M_{3} = 10 + 6 + 4 = 20 k g$

Acceleration of the system

$a = F / (M_{1} + M_{2} + M_{3}) = 40 / (20) = 2 m s^{- 2}$

Because string $T_{2}$ pulls 1 an 2 blocks
therefore
$T_{2} = (M_{1} + M_{2}) a = 16 \times 2 = 32 N$ .

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Given,Total force applied on the system = 40 N Total mass of the system= M1+M2+M3=10+6+4=20kg Acceleration of the system a=F/(M1+M2+M3)=40/(20)=2ms−2 Because string T2 pulls 1 an 2 blocksthereforeT2=(M1+M2)a=16×2=32 N.