  Question

Two blocks $$m_1\, =\, 4kg$$ and $$m_2\, =\, 2 kg$$, 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
4ms2,T=0  B
2ms2,T=5N  C
10ms2,T=10  D
15ms2,T=9  Solution

The correct option is A $$4\, ms^2, T \,= \,0$$$$m_{1}=4kg\\m_{2}=2kg\\\theta=37^{\circ}\\\mu_{1}=0.75\\\mu_{2}=0.25$$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. $$F=F_{1}-\mu_{1}m_{1}gcos\theta-\mu_{2}m_{2}gcos\theta\tag{1}$$Also, the net force down the incline is as given below. $$F=(m_{1}+m_{2})a\tag{2}$$Comparing $$(1)$$ and $$(2)$$, $$(m_{1}+m_{2})gsin\theta-(\mu_{1}m_{1}+\mu_{2}m_{2})gcos\theta=(m_{1}+m_{2})a$$Substituting the values,$$(4+2)10sin(37^{\circ})-(4\times0.75+2\times0.25)10cos(37^{\circ})=6a$$$$sin(37^{\circ})=0.6, cos(37^{\circ})=0.8$$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,$$T+16 = 16$$$$\therefore T = 0N$$ Physics

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