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Hint: First of all, we will calculate limiting frictional force by B on A and then we will find frictional force needed to move along with B if both are same then using its value we will find work done by using the formula $\text{work done }=\text{ Force}\times \text{displacement}$ and derive the final answer.

Formula used: ${{F}_{m}}=\mu mg$, ${{F}_{s}}=\dfrac{P}{{{m}_{1}}+{{m}_{2}}}\times {{m}_{2}}$

Complete step by step answer:

Now, as shown in figure in the question, we know that $100\ N$ force acts on B and drags it in one direction due to which frictional force is generated between B and A and thus, we will calculate the final frictional force acting on it.

First of all, we will denote $100\ N$ as force P, mass of object A as ${{m}_{A}}$, mass of object B as ${{m}_{B}}$. And for simplicity the direction in which object B is being dragged we will consider it as positive and opposite direction as negative.

Now, we know that the force P acts in positive direction so due to that frictional force always acts in direction opposite to it. So, the limiting frictional force by B on A will be in the opposite direction. The limiting frictional force is given by,

${{F}_{m}}=\mu mg$.

Now, as the frictional force acts on object A here mass will be ${{m}_{A}}$ so, ${{F}_{m}}=\mu {{m}_{A}}g$.

Now, the value of $\mu $ is 0.5 and $g$is taken as 9.81. So, substituting the values we will get,

${{F}_{m}}=\mu {{m}_{A}}g$

$\Rightarrow {{F}_{m}}=\left( 0.5 \right)\left( 10 \right)\left( 9.81 \right)=49.05\cong 50$

Hence, the limiting force is 50 N and as it acts in the opposite direction it will be $-50N$.

Now, we know that frictional force on mass 2 by mass 1 can be given by,

${{F}_{s}}=\dfrac{P}{{{m}_{1}}+{{m}_{2}}}\times {{m}_{2}}$

Now, we will substitute the values in this equation and then we will get

${{F}_{s}}=\dfrac{100}{10+10}\times 10=50$

Hence, the frictional force needed is also 50 N. As, limiting and frictional forces are same we will consider any one of them as force needed to drag and find the work done using formula,

$\text{work done }=\text{ Force}\times \text{displacement}$

$\Rightarrow \text{work done }=\text{ 50}\times \text{2=100}\ \text{J}$

Hence, work done is 100 J and as the frictional force is in the opposite direction the work done by that force will be $-100J$.

Thus, option (b) is correct.

Note: In this type of question, students must understand that here limiting and frictional forces are the same but if they are different in value then we have to take their difference to find the final force acting and in which direction it acts. The direction of force and frictional force should always be in opposite directions, students might take them in the same direction and then the answer will be wrong.

Formula used: ${{F}_{m}}=\mu mg$, ${{F}_{s}}=\dfrac{P}{{{m}_{1}}+{{m}_{2}}}\times {{m}_{2}}$

Complete step by step answer:

Now, as shown in figure in the question, we know that $100\ N$ force acts on B and drags it in one direction due to which frictional force is generated between B and A and thus, we will calculate the final frictional force acting on it.

First of all, we will denote $100\ N$ as force P, mass of object A as ${{m}_{A}}$, mass of object B as ${{m}_{B}}$. And for simplicity the direction in which object B is being dragged we will consider it as positive and opposite direction as negative.

Now, we know that the force P acts in positive direction so due to that frictional force always acts in direction opposite to it. So, the limiting frictional force by B on A will be in the opposite direction. The limiting frictional force is given by,

${{F}_{m}}=\mu mg$.

Now, as the frictional force acts on object A here mass will be ${{m}_{A}}$ so, ${{F}_{m}}=\mu {{m}_{A}}g$.

Now, the value of $\mu $ is 0.5 and $g$is taken as 9.81. So, substituting the values we will get,

${{F}_{m}}=\mu {{m}_{A}}g$

$\Rightarrow {{F}_{m}}=\left( 0.5 \right)\left( 10 \right)\left( 9.81 \right)=49.05\cong 50$

Hence, the limiting force is 50 N and as it acts in the opposite direction it will be $-50N$.

Now, we know that frictional force on mass 2 by mass 1 can be given by,

${{F}_{s}}=\dfrac{P}{{{m}_{1}}+{{m}_{2}}}\times {{m}_{2}}$

Now, we will substitute the values in this equation and then we will get

${{F}_{s}}=\dfrac{100}{10+10}\times 10=50$

Hence, the frictional force needed is also 50 N. As, limiting and frictional forces are same we will consider any one of them as force needed to drag and find the work done using formula,

$\text{work done }=\text{ Force}\times \text{displacement}$

$\Rightarrow \text{work done }=\text{ 50}\times \text{2=100}\ \text{J}$

Hence, work done is 100 J and as the frictional force is in the opposite direction the work done by that force will be $-100J$.

Thus, option (b) is correct.

Note: In this type of question, students must understand that here limiting and frictional forces are the same but if they are different in value then we have to take their difference to find the final force acting and in which direction it acts. The direction of force and frictional force should always be in opposite directions, students might take them in the same direction and then the answer will be wrong.

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