Question

Two blocks $A$ and $B$ of mass $2\text{kg}$ and $4\text{kg}$ respectively are placed one over the other. A horizontal force $F=2t$, which varies with time is applied on the upper block. If coefficiet of friction between blocks $A$ and $B$ is $0.5$ and horizontal surface is smooth. Then assuming $t$ is in seconds, the total time upto which both blocks will move together without slipping over each other is

• A. $5\text{sec}$
• B. $7.5\text{sec}$
• C. $10\text{sec}$
• D. $12.5\text{sec}$

Answer 1

The correct option is B $7.5\text{sec}$

The FBDs of the block is as shown
Given, external force applied $F=2t$ and blocks are moving together



If common maximum acceleration is $a_c$ then,
$F=(m_A+m_B)a_c$
$\Rightarrow 2t=(2+4)a_c$
$\Rightarrow t=3a_c$........(i)

Also, from the FBDs we have
$N_1=m_{A}g=2g=20N$
$N_2=N_1+4g=20+40=60N$
and, $f=f_{max}=\mu N_1=0.5\left(20\right)=10N$ at limiting case.

As there is no friction between block $B$ and surface so, for block $B$, we have
$f=m_B\times a_c$
$\Rightarrow 10=4a_c$
$\Rightarrow a_c=2.5{\text{ms}}^{-2}$
From equation (i), we get
$t=3a_c=3\times 2.5=7.5\text{sec}$