The Question is from neet-physics laws-of-motion , and it was created by the author P.NARASIMHA

Question

Two blocks of masses $m_1$ and $m_2$ are connected by a massless string passing over a smooth pulley.  The coefficient of kinetic friction between $m_1$ and the horizontal surface is $\mu_k$. If $m_2$ is descending with a constant acceleration $a$, what is the tension in the string?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Newton's Second Law of Motion
	**Related Topics**: Free Body Diagrams, Tension, Friction, Kinematics
	**Prerequisites**: Understanding of forces, Newton's Laws, and free-body diagrams
## SOLUTION PATHWAY
**Given Information**:
	* Masses: $m_1$, $m_2$
	* Coefficient of kinetic friction: $\mu_k$
	* Acceleration of $m_2$: $a$ (downward)
	* String is massless and pulley is smooth
**Method & Steps**:
	1. Draw free-body diagrams for both blocks.
		Reasoning: This helps visualize the forces acting on each block.
		Formula/Rule: N/A
	2. Apply Newton's Second Law to $m_1$ (horizontal motion).
		Working: The forces on $m_1$ are tension ($T$) pulling to the right and kinetic friction ($f_k$) pulling to the left. $f_k = \mu_k N = \mu_k m_1 g$. So, $T - f_k = m_1 a$. Therefore, $T - \mu_k m_1 g = m_1 a$.
		Key Point: The acceleration of $m_1$ is to the right and has the same magnitude as $m_2$ since they are connected by the string.
	3. Apply Newton's Second Law to $m_2$ (vertical motion).
		Working: The forces on $m_2$ are its weight ($m_2 g$) pulling downwards and tension ($T$) pulling upwards. So, $m_2 g - T = m_2 a$.
		Key Point: The acceleration of $m_2$ is downward.
	4. Solve for tension $T$ from the equations for $m_1$ and $m_2$.
		Working: From the equation for $m_1$: $T = m_1 a + \mu_k m_1 g = m_1(a + \mu_k g)$. From the equation for $m_2$: $T = m_2 g - m_2 a = m_2(g - a)$. Thus,  $T = m_1(a + \mu_k g) = m_2(g-a)$
		Result: $T = m_1(a + \mu_k g) = m_2(g-a)$
		Verification: The tension is consistent with both block's motion.
## OPTION ANALYSIS
[A] :x: This option only considers the weight of $m_2$, and doesn't account for acceleration or friction.
[B] :x: This option correctly calculates the tension based on $m_1$, but it does not equal $m_2(g-a)$.
[C] ✓: This option correctly provides the tension using both $m_1$ and $m_2$ equations and equates them to each other.
[D] :x: This option only considers the total mass and acceleration, but this doesn't account for tension.
## LEARNING AIDS
	**Common Mistakes**:
		1. [Error 1]: Incorrectly assigning directions to the acceleration in Newton's second law. [How to avoid]: Draw free body diagrams and define positive directions.
		2. [Error 2]: Forgetting to include the frictional force on $m_1$. [How to avoid]: Always check for all forces acting on each object.
	**Quick Tips**:
		1. [Helpful shortcut]:  When objects are connected by a string, they usually have the same magnitude of acceleration.
		2. [Memory aid]: Remember that tension always pulls along the string, and friction opposes motion.

P.NARASIMHA · Answer

$T = m_2g$

P.NARASIMHA · Answer

$T = m_1a + \mu_k m_1g$

P.NARASIMHA · Answer

The correct Option is "$T = m_1(a + \mu_k g) = m_2(g-a)$"

P.NARASIMHA · Answer

$T = (m_1 + m_2)a$

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