
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 be
A. \[\dfrac{{{m_2}g}}{{{m_1} + {m_2}}}\]
B. \[\dfrac{{{m_2}g}}{{{m_1} + {m_2}}}\]
C. \[\dfrac{{\mu \left( {{m_2} + {m_1}} \right)}}{{{m_1} + {m_2}}}g\]
D. \[\left[ {\dfrac{{{m_2} - \mu {m_1}}}{{{m_1} + {m_2}}}} \right]g\]

Answer
476.7k+ 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,

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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
