Question

Two blocks $m_1=4kg$ and $m_2=2kg$, connected by a weightless rod on a plane having inclination of ${37}^{\circ }$ as shown in figure. 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 (take sin37 = 3/5)

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

Answer 1

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

$m_1=4kg{m}_2=2kg\theta ={37}^{\circ }{\mu }_1=0.75{\mu }_2=0.25$
Let $a$ be the common acceleration of the blocks and $T$ be the tension in the connecting rod.
Force down the incline $=F_1=m_{1}gsin\theta +m_{2}gsin\theta =(m_1+m_2)gsin\theta$
Now, normal reaction on $m_1=R_1=m_{1}gcos\theta$
Frictional force on $m_1=f_1={\mu }_1{R}_1={\mu }_1{m}_{1}gcos\theta$
Also, normal reaction on $m_2=R_2=m_{2}gcos\theta$
Frictional force on $m_2=f_2={\mu }_2{R}_2={\mu }_2{m}_{2}gcos\theta$
Net force down the incline is as given below. $\begin{array}{}F=F_1-{\mu }_1{m}_{1}gcos\theta -{\mu }_2{m}_{2}gcos\theta & \text{(1)}\end{array}$
Also, the net force down the incline is as given below. $\begin{array}{}F=(m_1+m_2)a& \text{(2)}\end{array}$
Comparing $\left(1\right)$ and $\left(2\right)$,
$(m_1+m_2)gsin\theta -({\mu }_1{m}_1+{\mu }_2{m}_2)gcos\theta =(m_1+m_2)a$
Substituting the values,
$(4+2)10sin\left({37}^{\circ }\right)-(4\times 0.75+2\times 0.25)10cos\left({37}^{\circ }\right)=6a$
$sin\left({37}^{\circ }\right)=0.6,cos\left({37}^{\circ }\right)=0.8$
Substituting and solving,
$24=6a$
$\therefore a=4m/s^2$
To find T,
Equation of motion for block $m_1$ is as given below.
$T+m_{1}gsin\theta -{\mu }_1{m}_{1}gcos\theta =m_{1}a$
Substituting,
$T+16=16$
$\therefore T=0N$