1. Tardigrade
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  4. Two boys of masses 10 kg and 8 kg are moving along a vertical rope, the former climbing up with the acceleration of 2 m s- 2 while the later coming down with a uniform velocity of 2 m s- 1 . Then tension in the rope at the fixed support is (Take g = 10 m s- 2 )

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Q. Two boys of masses $10 \, kg$ and $8 \, kg$ are moving along a vertical rope, the former climbing up with the acceleration of $2 \, m \, s^{- 2}$ while the later coming down with a uniform velocity of $2 \, m \, s^{- 1}$ . Then tension in the rope at the fixed support is (Take $g \, = \, 10 \, m \, s^{- 2}$ )

NTA AbhyasNTA Abhyas 2020Laws of Motion

Solution:

Since, $m_{2}$ moves with constant velocity.
$\therefore $ $f_{2}=m_{2}g$
$f_{2}=8 \, \times 10=80N$
Since, block $m_{1}$ moves with acceleration

Solution
$a=2ms^{- 2}$ in the upward direction.
$\therefore $ $f_{1}-m_{1}g=m_{1}a$
$\therefore $ $f_{1}=m_{1}g+m_{1}a$
$=10 \, \times 10+10\times 2$
$=120 \, N$
For equilibrium of rope,

Solution
$T=f_{1}+f_{2}$
$=\left(120 + 80\right)N$
$=200N$