NEET 2026 Physics: Mastering the Laws of Motion Friction Trap Question

Hello future doctors!

If you're scoring below 350 in NEET, especially in Physics, you might feel like certain questions are designed just to trip you up. And honestly, some are! But don't worry, every 'trap' question has a logical explanation, and once you see it, you'll wonder why it ever seemed hard. Today, we're diving deep into a classic NEET Physics trap from the Laws of Motion — a type of question that appeared in NEET 2024 (and similar variations in previous years). We'll dissect it, understand exactly why students get it wrong, and build your confidence to conquer it.

Physics is all about understanding the underlying principles. Sometimes, the difficulty isn't in complex math, but in correctly applying basic concepts. This question pattern tests your fundamental understanding of Newton's Laws, Free Body Diagrams, and the ever-tricky concept of friction.

The Trap Question: Laws of Motion with Friction

This type of question appeared in NEET 2024 — here is exactly why students get it wrong.

Question: A block of mass M₁ = 5 kg is placed on a rough horizontal surface. It is connected by a light inextensible string passing over a frictionless pulley to another block of mass M₂ = 3 kg hanging vertically. If the coefficient of kinetic friction between M₁ and the surface is μk = 0.2, and g = 10 m/s², what is the acceleration of the system?

(A) 2.5 m/s²

(B) 3.75 m/s²

(C) 3.0 m/s²

(D) 1.25 m/s²

Why Students Get This Wrong: The Traps Laid Out

Let's unpack the common pitfalls. These aren't just random wrong answers; they're carefully crafted to exploit predictable mistakes.

• TRAP (B) 3.75 m/s² — Ignoring Friction: Students choose this because they might forget to account for friction, or they might think the term 'frictionless pulley' means no friction anywhere. They'd calculate the net pulling force as M₂g and the total mass as M₁+M₂, leading to:

  a = (M₂g) / (M₁+M₂) = (3 kg × 10 m/s²) / (5 kg + 3 kg) = 30 N / 8 kg = 3.75 m/s². This is a classic conceptual error, assuming an ideal scenario when friction is clearly mentioned.

• TRAP (C) 3.0 m/s² — Incorrect Normal Force for Friction: This is a sneaky one! Students know friction is μN, but they might mistakenly use M₂g as the normal force (N) for the block on the surface, instead of M₁g. Remember, the normal force on M₁ is only due to its own weight on the horizontal surface. So, if they calculate friction as:

  f_k = μk × M₂g = 0.2 × 3 kg × 10 m/s² = 6 N (This is WRONG!)

  Then, the acceleration would be a = (M₂g - f_k) / (M₁+M₂) = (30 N - 6 N) / 8 kg = 24 N / 8 kg = 3.0 m/s². This shows a common misunderstanding of how normal force applies in different parts of a system.

• TRAP (D) 1.25 m/s² — Overestimating Friction (or other calculation errors): This option often comes from miscalculating the friction force, or making a larger arithmetic error. For instance, if a student mistakenly used a coefficient of friction μk = 0.4 (perhaps a typo in their notes or a misreading of the question), the friction force would be:

  f_k = 0.4 × M₁g = 0.4 × 5 kg × 10 m/s² = 20 N.

  Then, a = (M₂g - f_k) / (M₁+M₂) = (30 N - 20 N) / 8 kg = 10 N / 8 kg = 1.25 m/s². This highlights the importance of careful reading and calculation.

The Correct Approach: Back to Basics!

To solve this, we need to apply Newton's Second Law (F_net = ma) to each block separately, using Free Body Diagrams (FBDs).

Step 1: Draw Free Body Diagrams (FBDs)

This is the most crucial step. If your FBDs are wrong, your equations will be wrong.

• For M₁ (on the rough horizontal surface):

  • Weight (M₁g): Acts downwards.

  • Normal Force (N): Acts upwards, perpendicular to the surface. Since it's a horizontal surface and no other vertical forces, N = M₁g.

  • Tension (T): Acts horizontally to the right (pulling force).

  • Kinetic Friction (f_k): Acts horizontally to the left (opposing motion). f_k = μk × N = μk × M₁g.

• For M₂ (hanging vertically):

  • Weight (M₂g): Acts downwards.

  • Tension (T): Acts upwards (pulling force from the string).

Step 2: Write Newton's Second Law Equations

Assume the system accelerates to the right (M₁ moves right, M₂ moves down) with acceleration 'a'.

• For M₁ (horizontal motion):
  The net force in the direction of motion is (T - f_k).
  T - f_k = M₁a

• For M₂ (vertical motion):
  The net force in the direction of motion is (M₂g - T).
  M₂g - T = M₂a

Step 3: Calculate the Friction Force (f_k)

First, find the Normal Force (N) for M₁:

N = M₁g = 5 kg × 10 m/s² = 50 N.

Now, calculate kinetic friction:

f_k = μk × N = 0.2 × 50 N = 10 N.

Step 4: Substitute and Solve the System of Equations

Our equations are:

1. T - 10 = 5a

2. 30 - T = 3a

We have two equations and two unknowns (T and a). The easiest way to solve is to add the two equations to eliminate T:

(T - 10) + (30 - T) = 5a + 3a

20 = 8a

a = 20 / 8 = 2.5 m/s²

Thus, the correct answer is Option (A) 2.5 m/s².

Memory Shortcut: The 'Net Pulling Force' Method

For simple systems like this, you can often use a shortcut once you've correctly identified all forces:

a = (Net Pulling Force - Total Opposing Force) / Total Mass of System

Let's apply it:

• Net Pulling Force: The force trying to move the system is the weight of the hanging mass, M₂g = 3 kg × 10 m/s² = 30 N.

• Total Opposing Force: The only force opposing the motion here is kinetic friction on M₁, which is f_k = μk M₁g = 0.2 × 5 kg × 10 m/s² = 10 N.

• Total Mass of System: M₁ + M₂ = 5 kg + 3 kg = 8 kg.

So, a = (30 N - 10 N) / 8 kg = 20 N / 8 kg = 2.5 m/s².

This shortcut works perfectly, but only if you are absolutely clear on which forces are 'pulling' and which are 'opposing', and how to calculate them correctly (especially friction!). If you're ever unsure, always fall back on drawing FBDs and writing separate equations.

⚠️ Most Dangerous Mistake ⚠️

Similar PYQ Watch List (NEET 2024-2026)

This pattern is a perennial favorite. Keep an eye out for variations:

1. Block on an Inclined Plane with Friction: Similar concepts, but the normal force (N) will be M g cosθ and the component of gravity pulling it down is M g sinθ. Friction will still oppose motion.

2. Two Blocks Connected, One Pushed by External Force: Instead of a hanging mass, an external force pushes one block, and the string connects it to another. Here, you'll analyze forces on each block and the tension in the string.

3. System with Two Pulleys or Multiple Blocks: More complex FBDs, but the core principle of F_net = ma for each part remains the same. The key is to be consistent with directions and forces.

NEET Probability Tag: High

Questions involving Newton's Laws, friction, and pulley systems appear frequently in NEET. Mastering this type of problem is essential for securing marks in Physics.

Remember, every mistake is a learning opportunity. Don't feel demoralized if you got this wrong initially. The goal is to understand *why* you made the mistake and learn to avoid it next time. Practice is key!

Don't struggle alone – TheRishiPath app has detailed video solutions and practice questions for every concept like this. Download it today! It's like having a top NEET teacher in your pocket, guiding you through every tricky problem.

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