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Byju's Answer

Standard XII

Physics

Elastic Potential Energy

Three blocks ...

Question

Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless string as shown on a frictionless table. They are 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

Total mass of system = $20 k g$
Force = $40 N$
Accelration is:
$A = \frac{f}{m}$

= $\frac{40}{20}$

= $2 \frac{m}{s^{- 2}}$
Because string $T_{2}$ pulls $1$ and $2$ blocks

Therefore $T_{2} = (m_{1} + m_{2}) \times (a)$

= $32 N$

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Three blocks of masses $m_{1}$ , $m_{2}$ and $m_{3}$ are connected by massless strings as shown in the figure and placed on a frictionless table. A force of $40 N$ pulls the system. Then calculate the value of $T$ (in $N$ ) if $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$ .

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Three blocks of masses m1,m2 and m3 are connected by massless string as shown on a frictionless table. They are pulled with a force F=40N. If m1=10kg, m2=6kg and m3=4kg, then tension T2 will be:

Total mass of system = 20kgForce = 40NAccelration is:A=fm = 4020 = 2ms−2Because string T2 pulls 1 and 2 blocksTherefore T2=(m1+m2)×(a) = 32N

Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless string as shown on a frictionless table. They are 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:

Total mass of system = $20 k g$
Force = $40 N$
Accelration is:
$A = \frac{f}{m}$

= $\frac{40}{20}$

= $2 \frac{m}{s^{- 2}}$
Because string $T_{2}$ pulls $1$ and $2$ blocks

Therefore $T_{2} = (m_{1} + m_{2}) \times (a)$

= $32 N$