Question

Two blocks of mass $m_1=4\text{kg}$ and $m_2=2\text{kg}$, connected by a weightless rod on a plane having inclination of ${37}^{\circ }$ is shown in figure. If the coefficient of dynamic friction is $\mu =0.25$, then the common acceleration of the two blocks and the tension $\left(T\right)$ in the rod are

• A. $4{\text{m/s}}^2,T=0\text{N}$
• B. $2{\text{m/s}}^2,T=5\text{N}$
• C. $10{\text{m/s}}^2,T=10\text{N}$
• D. $15{\text{m/s}}^2,T=9\text{N}$

Answer 1

The correct option is A $4{\text{m/s}}^2,T=0\text{N}$

Both the blocks are connected with the rod so both the blocks will moves with common acceleration. Simultaneously considering blocks and rod as a system.
F.B.D. of system,


From the F.B.D,
$N_2=m_{2}g\cos \theta$ and $N_1=m_{1}g\cos \theta$
$f_2=\mu N_2$ and $f_1=\mu N_1$
$\therefore f_2=\mu m_{2}g\cos \theta$
and $f_1=\mu m_{1}g\cos \theta$
Total frictional force on the system,
$f=f_1+f_2=\mu g\cos \theta (m_1+m_2)$

Assuming $a$ is the common acceleration along the surface.

$(m_1+m_2)g\sin \theta -(m_1+m_2)\mu g\cos \theta =(m_1+m_2)a$
$\Rightarrow 10\times \sin {37}^{\circ }-0.25\times 10\times \cos {37}^{\circ }=a$
$\Rightarrow 10\times \frac{3}{5}-0.25\times 10\times \frac{4}{5}=a$
$\Rightarrow a=4{\text{m/s}}^2$

Now for tension take $m_2$ block as a system F.B.D. $m_2$ Block

$f=\mu m_{2}g\cos \theta$
Therefore,
$m_{2}g\sin \theta -f-T=m_{2}a$
$\Rightarrow m_{2}g\sin \theta -\mu m_{2}g\cos \theta -T=m_{2}a$
$\Rightarrow 2\times 10\times \frac{3}{5}-0.25\times 2\times 10\times \frac{4}{5}-T=2\times 4$
$\Rightarrow 12-4-T=8$
$T=0\text{N}$