Question

A wooden block $A$ of mass $10\text{kg}$ is placed on another identical block $B$ of mass $15\text{kg}$ lying on a horizontal wooden table as shown below. It is found that when a force of $20\text{N}$ is applied on the block $A$, it just begins to slide over the surface of the block $B$. When a body of mass $7.5\text{kg}$ is connected to the block $B$ by a string and pulley arrangement so that both the blocks $A$ and $B$ just move together, the co-efficient of friction between the block $B$ and surface will be:-
(Take $g=10{\text{ms}}^{-2}$)

• A. $0.267$
• B. $0.250$
• C. $0.220$
• D. $0.350$

Answer 1

The correct option is C $0.220$

Here, $M_A=10\text{kg},M_B=15\text{kg}$.
As on applying a force of $20\text{N}$ on the block $A$, it just begins to slide over the surface of the block $B$, that means applied force is equal to the limiting friction between the two blocks. i.e $F_{max}=20\text{N}$


If $\mu$ is the co-efficient of friction between the block surfaces, then
$F_{max}=\mu \times M_A\times g$
$\Rightarrow \mu =\frac{F_{max}}{M_{A}g}=\frac{20}{10\times 10}=0.2$
Let ${\mu }^{\prime }$ be the coefficient of friction between the surface and block $B$.
Now, free body diagram for block $B$ and $7.5\text{kg}$ mass:


Since the system just starts sliding on the table when a body of mass $7.5\text{kg}$ is connected to the block $B$.
From FBD of mass $7.5\text{kg}$,
$T=7.5g=7.5\times 10=75\text{N}$
From FBD of body $B$,
$T=\mu M_{A}g+{\mu }^{\prime }(M_A+M_B)g$
$\Rightarrow 75=(0.2\times 10\times 10)+({\mu }^{\prime }\times 25\times 10)$
$\Rightarrow {\mu }^{\prime }=\frac{55}{250}\approx 0.22$
Hence, option $\left(c\right)$ is the correct answer.