Question

In the diagram shown below the coefficient of friction between $2\text{kg}$ and $4\text{kg}$ block is $0.1$ and between $4\text{kg}$ and ground is $0.2$. Find the force of friction on $2\text{kg}$ block at time $t=0.5$ second.

• A. $2\text{N towards left}$
• B. $1\text{N towards right}$
• C. $1\text{N towards left}$
• D. $0\text{N}$

Answer 1

The correct option is B $1\text{N towards right}$

The FBD of the blocks are as shown


From the FBD we have
$N_1=2g=20\text{N}$
Maximum available value of friction between $2\text{kg}$ and $4\text{kg}$ block is
$(f_1{)}_{\text{max}}=\mu N_1=0.1\times 20=2\text{N}$

and, $N_2=N_1+4g=20+40=60\text{N}$
Similarly, maximum available value of friction between $4\text{kg}$ and ground is
$(f_2{)}_{\text{max}}=\mu N_2=0.2\times 60=12\text{N}$

Now, it is given that $F=2t$ towards left
So, at $t=0.5\text{s},F=2\times 0.5=1\text{N}$
$\because$ $F_{@t=0.5s}<(f_1{)}_{\text{max}}$, the whole system is at rest, the tension will not develop in the string and the friction force acting on $2kg$ block will be $1\text{N}$ towards right.