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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)

293089_13b3d9c4521347eda492c0b4ef557f9d.png


A
4ms2,T=0
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B
2ms2,T=5N
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C
10ms2,T=10
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D
15ms2,T=9
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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,
$$T+16 = 16$$
$$\therefore T = 0N$$
282100_293089_ans_9bf2f4ed00ba488b8bf57b57c627abd7.png

Physics

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