Question

Two blocks of mass $m_1$ and $m_2$ are connected by a weightless rod on a plane having inclination of ${37}^{\circ }$. The coefficients of dynamic friction of $m_1$ and $m_2$ with the inclined plane are $\mu =0.25$. Then the common acceleration of the two blocks and the tension in the rod are

• A. $4{\text{m/s}}^2,T=0$
• 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 $4{\text{m/s}}^2,T=0$

From FBD of $m_2$


$m_{2}g\sin \alpha -\mu m_{2}g\cos \alpha -T=m_{2}a$ ... (1)
From FBD of $m_1$
$m_{1}g\sin \alpha +T-\mu m_{1}g\cos \alpha =m_{1}a$ ... (2)
Solving (1) and (2) for the value of $T$,
$m_1{m}_{2}g\sin \alpha -\mu m_1{m}_{2}g\cos \alpha -m_{1}T=m_1{m}_{2}g\sin \alpha +m_{2}T-\mu m_1{m}_{2}g\cos \alpha$
$\Rightarrow T(m_1+m_2)=0$
$\therefore T=0$ ... (3)
Substituting the value of (3) in either (1) or (2) we get
$a=g\sin \alpha -\mu g\cos \alpha$
$a=10\sin {37}^{\circ }-(0.25\times 10\cos {37}^{\circ })$
$a=4{\text{m/s}}^2$