Arrangement of two block system is as shown. The net force acting on 1 kg and 2 kg blocks are (assuming the surfaces to be frictionless) respectively

Question

Arrangement of two block system is as shown. The net force acting on 1 kg and 2 kg blocks are (assuming the surfaces to be frictionless) respectively (A) 4 N , 8 N (B) 1 N , 2 N (C) 2 N , 4 N (D) 3 N , 6 N

Tardigrade Admin · Accepted Answer

acceleration of the system =(F text ext /M text Total ) =(6/3)=2 m / s 2

for 2 kg block N=2(2)=4 N

for 1 kg block F text net =6-4=2 N

Q. Arrangement of two block system is as shown. The net force acting on $1 k g$ and $2 k g$ blocks are (assuming the surfaces to be frictionless) respectively

4 N , 8 N

1 N , 2 N

2 N , 4 N

3 N , 6 N

acceleration of the system $= \frac{F _{ext}}{M _{Total}}$
$= \frac{6}{3} = 2 m / s^{2}$

for $2 k g$ block
$N = 2 (2) = 4 N$

for $1 k g$ block
$F_{net} = 6 - 4 = 2 N$

Q. Arrangement of two block system is as shown. The net force acting on 1kg and 2kg blocks are (assuming the surfaces to be frictionless) respectively