Question

In the given figure, the blocks of mass $2\text{kg}$ and $3\text{kg}$ are placed one over the other as shown. The surface are rough with coefficient of friction ${\mu }_1=0.2,{\mu }_2=0.06$. A force $F=0.5t$ (where ${}^{\prime }{t}^{\prime }$ is time in sec) is applied on upper block in the direction shown. The frictional force acting between $3\text{kg}$ block and ground w.r.t. time will vary as
Take ($g=10{\text{m/s}}^2$)

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is C

The FBDs of the blocks are as shown

From the FBDs we have
$N_1=2g=20N$
$N_2=N_1+3g=20+30=50N$
$f_{\text{1 max}}=\mu N_1=\left(0.2\right)\left(20\right)=4\text{N}$
$f_{\text{2 max}}=\mu N_2=\left(0.06\right)\left(50\right)=3\text{N}$

The two blocks will start moving when $F=f_{\text{2 max}}\Rightarrow 0.5t=3\Rightarrow t=6\text{sec}$
Till then, friction is static in nature and increases with increase in applied force.
Then $f_2$ will become kinetic in nature and remains constant.