Question

Two blocks of masses $2kg$ and $4kg$ are connected by a light string and kept on horizontal surface. A force of $16N$ is acted on 4kg block horizontally as shown in figure. Besides it is given that coefficient of friction between 4~kg and ground is 0.3 and between $2kg$ block and ground is 0.6. Then frictional force between $2kg$ block and ground is

• A. $12N$
• B. $4N$
• C. $6N$
• D. zero

Answer 1

The correct option is B $4N$

Let's first find the overall acceleration of system ($2kg$ and $4kg$ blocks)
Limiting friction can act between the surfaces of blocks are
1. $(F_s{)}_{max}$ between mass $2kg$ block and ground is $0.6\times 2\times 10=12N$.
2. Similarly, $(F_s{)}_{max}$ between mass $4kg$ block and ground is $0.3\times 4\times 10=12N$.
Total limiting friction which can be act between blocks and ground is $12N+12N=24N$ which is greater than the force applied $16N$. Hence the blocks will not move. So the acceleration of system is zero and it is in equilibrium.
Let's consider the FBD of $4kg$ block and $2kg$ block,


From FBD of $4kg$ we can calculate the tension in string which is equal to $T=16-12=4N$
Hence from the FBD of $2kg$ block,
$F_s=4N$