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 )

Question

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 ) (A) 200N (B) 120N (C) 180N (D) 160N

Tardigrade Admin · Accepted Answer

Since, m2 moves with constant velocity. ∴ f2=m2g f2=8 × 10=80N Since, block m1 moves with acceleration Solution

a=2ms- 2 in the upward direction. ∴ f1-m1g=m1a ∴ f1=m1g+m1a =10 × 10+10× 2 =120 N For equilibrium of rope, Solution

T=f1+f2 =(120 + 80)N =200N

Q. Two boys of masses $10 k g$ and $8 k g$ 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 $g = 10 m s^{- 2}$ )

$200 N$

$120 N$

$180 N$

$160 N$

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

200N

120N

180N

160N

Since, m2​ moves with constant velocity. ∴ f2​=m2​g f2​=8×10=80N Since, block m1​ moves with acceleration a=2ms−2 in the upward direction. ∴ f1​−m1​g=m1​a ∴ f1​=m1​g+m1​a =10×10+10×2 =120N For equilibrium of rope, T=f1​+f2​ =(120+80)N =200N

Q. Two boys of masses $10 k g$ and $8 k g$ 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 $g = 10 m s^{- 2}$ )

$200 N$

$120 N$

$180 N$

$160 N$