Three identical blocks of masses m=2 kg are drawn by a force F=10.2 N with an acceleration of 0.6 ms -2 on a frictionless surface, then what is the tension (in N ) in the string between the blocks B and C ?

Question

Three identical blocks of masses m=2 kg are drawn by a force F=10.2 N with an acceleration of 0.6 ms -2 on a frictionless surface, then what is the tension (in N ) in the string between the blocks B and C ? (A) 9.2 (B) 7.8 (C) 4 (D) 9.8

Tardigrade Admin · Accepted Answer

The system of masses is shown below. From the figure.

F - T_{1} = ma

and

T_{1} - T_{2} = ma

Eq. (1) gives

10.2 - T_{1} = 2 \times 0.6

\Rightarrow T_{1} = 10.2 - 1.2 = 9 N

Again from Eq. (2), we get

9 - T_{2} = 2 \times 0.6

\Rightarrow T_{2} = 9 - 1.2 = 7.8 N

Q. Three identical blocks of masses m=2kg are drawn by a force F=10.2N with an acceleration of 0.6ms−2 on a frictionless surface, then what is the tension (in N ) in the string between the blocks B and C ?

9.2

7.8

4

9.8

The system of masses is shown below. From the figure. F−T1​=ma and T1​−T2​=ma Eq. (1) gives 10.2−T1​=2×0.6 ⇒T1​=10.2−1.2=9N Again from Eq. (2), we get 9−T2​=2×0.6 ⇒T2​=9−1.2=7.8N

Q. Three identical blocks of masses $m = 2 k g$ are drawn by a force $F = 10.2 N$ with an acceleration of $0.6 m s^{- 2}$ on a frictionless surface, then what is the tension (in $N$ ) in the string between the blocks $B$ and $C$ ?

The system of masses is shown below. From the figure.

$F - T_{1} = ma$
and $T_{1} - T_{2} = ma$
Eq. (1) gives
$10.2 - T_{1} = 2 \times 0.6$
$\Rightarrow T_{1} = 10.2 - 1.2 = 9 N$
Again from Eq. (2), we get
$9 - T_{2} = 2 \times 0.6$
$\Rightarrow T_{2} = 9 - 1.2 = 7.8 N$