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# A block of mass ${m_1}$ is resting on a rough horizontal plane, the coefficient of the kinetic friction between the block and the surface is $\mu$. If ${m_1}$ is connected to another mass ${m_2}$ with the help of string and pulley as shown in the figure, then the common acceleration when the system is released from the rest will beA. $\dfrac{{{m_2}g}}{{{m_1} + {m_2}}}$B. $\dfrac{{{m_2}g}}{{{m_1} + {m_2}}}$C. $\dfrac{{\mu \left( {{m_2} + {m_1}} \right)}}{{{m_1} + {m_2}}}g$D. $\left[ {\dfrac{{{m_2} - \mu {m_1}}}{{{m_1} + {m_2}}}} \right]g$

Last updated date: 13th Jun 2024
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Hint: Apply Newton’s second law of motion to both the masses and rearrange the equations you got to determine the effective acceleration of the system.

Draw a free body diagram to denote the forces acting on the above system as follows,

In the above free body diagram, ${F_r}$ is the frictional force acting on the mass ${m_1}$, T is the tension in the string, and ${m_2}g$ is the weight of ${m_2}$.
Apply Newton’s second law of motion on the mass ${m_1}$ as follows,
${F_{net}} = {m_1}a$
$\Rightarrow T - {F_r} = {m_1}a$ …… (1)
We know that the frictional force acting on the mass ${m_1}$ is given by the equation,
${F_r} = \mu {m_1}g$
Therefore, equation (1) becomes,
$\Rightarrow T - \mu {m_1}g = {m_1}a$ …… (2)
Apply Newton’s second law of motion on the mass ${m_2}$ as follows,
$T - {m_2}g = {m_2}\left( { - a} \right)$
$\Rightarrow T - {m_2}g = - {m_2}a$ …… (3)
Here, the negative sign of the acceleration implies that the mass is accelerated in the downward direction.
Subtract equation (2) from equation (3).
$T - {m_2}g - \left( {T - \mu {m_1}g} \right) = - {m_2}a - {m_1}a$
$\Rightarrow {m_2}g - \mu {m_1}g = \left( {{m_1} + {m_2}} \right)a$
$\therefore a = \dfrac{{\left( {{m_2} - \mu {m_1}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}g$
So, the correct answer is “Option D”.

Note:
Always specify the direction of the forces and the direction of the acceleration of the body. In this problem, the direction of the mass ${m_1}$ is towards right and hence taken positive while the direction of the acceleration of the mass ${m_2}$ is downwards, hence taken as negative.