Question

Two blocks $m_1=4kg$ and $m_2=2kg$ are connected by a weightess rod on a plane having inclination of ${37}^{\circ }$. The coefficient of dynamic friction for blocks $m_1$ and $m_2$ with the inclined plane is $\mu =0.2$. Assume
${\mu }_k={\mu }_s=0.2$ and g=10 ${ms}^{-2}$. Then the common acceleration of the two blocks and the tension in the rod will be:

• A. $a=4.4m/s^2,T=0$
• B. $a=2m/s^2,T=5N$
• C. $a=10m/s^2,T=10N$
• D. $a=15m/s^2,T=9N$

Answer 1

The correct option is A $a=4.4m/s^2,T=0$



From FBD of $m_2$
$m_{2}gsin\alpha -\mu m_{2}gcos\alpha -T=m_{2}a----\left(1\right)$
From FBD of $m_1$
$m_{1}gsin\alpha +T-\mu m_{1}gcos\alpha =m_{1}a----\left(2\right)$
Solving (1) and (2) for the value of Tension $T$
$\Rightarrow T(m_1+m_2)=0$
$\therefore T=0---\left(3\right)$
Substituting the value of (3) in either (1) on (2) we get
$a=gsin\alpha -\mu gcosa$
$=gsin{37}^{\circ }-\mu gcos{37}^{\circ }$
$=4.4m/s^2$