Question

Co-efficient of friction between $5\text{kg}$ and $10\text{kg}$ blocks is $0.5$. If friction between them is $20\text{N}$. What is the value of force being applied on $5\text{kg}$. Assume the floor is friction less.

• A. $10\text{N}$
• B. $20\text{N}$
• C. $30\text{N}$
• D. $40\text{N}$

Answer 1

The correct option is C $30\text{N}$

FBD of $5\text{kg}$ block is given by,


Limiting friction, $f_l=0.5\times 50=25\text{N}$

As friction is $20\text{N}$ which is less than limiting friction $25\text{N}$, this means the friction is static in nature.

Therefore, $5\text{kg}$ and $10\text{kg}$ blocks will have no relative motion and acceleration of both will be same.

So, $F-20=5a$ ------ (1)

FBD of $10\text{kg}$ block is given by,


$f=10a$

$20=10a$

$\Rightarrow a=2\text{m}/{\text{s}}^2$ ----- (2)

From $\left(1\right)$ and $\left(2\right)$

$F-20=5\times 2$

$\therefore F=30\text{N}$

Hence, option (c) is the correct answer.