Answer
414.9k+ views
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.
Complete step by step answer:
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.
Complete step by step answer:
Draw a free body diagram to denote the forces acting on the above system as follows,
![seo images](https://www.vedantu.com/question-sets/2b116bf0-156f-47ca-b490-d199cf5f2d0a7200075774872827757.png)
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.
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