Physics - Mechanics and Properties of Matter - How to Solve: Newton’s Laws of Motion (Free Body Diagram, Pseudo Force, Tension, Spring Force) – NEET UG Guide

How to Solve: Newton’s Laws of Motion (Free Body Diagram, Pseudo Force, Tension, Spring Force) – NEET UG Guide

Introduction

Mastering Newton’s Laws unlocks 8–10 marks in NEET Physics—enough to push you from the 50th to the 70th percentile. Whether it’s a block on an incline, a pulley system, or a car accelerating, these problems test your ability to draw forces, apply Newton’s laws, and solve for unknowns—skills that also apply to circular motion, gravitation, and even fluid dynamics.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand: 1. Vector addition and resolution (breaking forces into x and y components). 2. Newton’s three laws of motion (inertia, F=ma, action-reaction). 3. Basic trigonometry (sin, cos, tan for inclined planes).

If any of these are shaky, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Newton’s Second Law (Vector Form)
   Fₙₑₜ = ma
2. Fₙₑₜ = Net force (vector sum of all forces) [N]
3. m = Mass of the object [kg]

4. a = Acceleration of the object [m/s²]
   MEMORISE THIS – This is the core equation for all Newton’s laws problems.

5. Pseudo Force (Non-Inertial Frame)
   Fₚₛₑᵤdₒ = -ma₀

6. Fₚₛₑᵤdₒ = Pseudo force [N]
7. a₀ = Acceleration of the non-inertial frame (e.g., accelerating car) [m/s²]

8. Direction: Opposite to the frame’s acceleration.
   MEMORISE THIS – Only applies in accelerating reference frames.

9. Hooke’s Law (Spring Force)
   Fₛ = -kx

10. Fₛ = Spring force [N]
11. k = Spring constant [N/m]
12. x = Displacement from equilibrium [m]

13. Negative sign = Force is opposite to displacement.
    MEMORISE THIS – Given on NEET formula sheet, but understand the sign.

14. Tension in a String

15. Tension is uniform in a massless, inextensible string.
16. If the string has mass, tension varies along its length.
17. MEMORISE: For massless strings, T is the same everywhere.

STEP-BY-STEP METHOD

Follow these 5 steps for every Newton’s laws problem:

Step 1: Identify the System & Frame of Reference

• Is the frame inertial (non-accelerating) or non-inertial (accelerating)?
• Inertial (e.g., ground, constant velocity): No pseudo force.
• Non-inertial (e.g., accelerating car, lift): Add pseudo force opposite to acceleration.
• Example: A block in an accelerating lift → non-inertial frame → add pseudo force.

Step 2: Draw the Free Body Diagram (FBD)

• Isolate the object (draw it as a dot or box).
• Draw all forces acting on it:
• Gravity (mg) – Always downward.
• Normal force (N) – Perpendicular to surface.
• Tension (T) – Away from the object (along the string).
• Friction (f) – Opposite to motion (or tendency to move).
• Spring force (Fₛ) – Opposite to displacement (if spring is stretched/compressed).
• Pseudo force (Fₚₛₑᵤdₒ) – Only in non-inertial frames, opposite to acceleration.
• Label all forces clearly (e.g., T₁, T₂, N₁, N₂ if multiple).

Step 3: Resolve Forces into Components

• Choose x and y axes (usually along the surface and perpendicular to it).
• Break forces into x and y components using trigonometry:
• Inclined plane? Resolve mg into mg sinθ (parallel) and mg cosθ (perpendicular).
• Pulley system? Align axes along the string.
• Write equations for each axis:
• ΣFₓ = maₓ (sum of forces in x-direction = mass × acceleration in x)
• ΣFᵧ = maᵧ (sum of forces in y-direction = mass × acceleration in y)

Step 4: Apply Newton’s Second Law (Fₙₑₜ = ma)

• For each axis, write the equation:
• If acceleration = 0 (equilibrium): ΣF = 0.
• If accelerating: ΣF = ma.
• Solve for unknowns (e.g., tension, normal force, acceleration).

Step 5: Check Units & Reasonableness

• Units: All forces in Newtons (N), mass in kg, acceleration in m/s².
• Reasonableness:
• Is the acceleration too high (e.g., > 10 m/s² for everyday objects)?
• Is the tension greater than the weight (possible if accelerating upward)?
• Does the normal force make sense (e.g., N < mg on an incline)?

WORKED EXAMPLES

Example 1 – Basic (Block on Horizontal Surface)

Problem: A 5 kg block is pulled by a 20 N horizontal force on a frictionless surface. Find its acceleration.

Step 1: System & Frame

• Inertial frame (ground, no acceleration of frame).
• No pseudo force needed.

Step 2: Free Body Diagram (FBD)

• Forces:
• mg = 5 × 9.8 = 49 N (downward).
• N (upward, from surface).
• F = 20 N (horizontal, applied force).
• No friction (given).

Step 3: Resolve Forces

• x-axis: F = 20 N (only horizontal force).
• y-axis: N = mg = 49 N (no vertical acceleration).

Step 4: Apply F = ma

• x-axis: ΣFₓ = ma → 20 = 5a → a = 4 m/s².
• y-axis: ΣFᵧ = 0 → N = mg (already balanced).

Step 5: Check

• Units: 20 N = 5 kg × 4 m/s² → Correct.
• Reasonable: 4 m/s² is a moderate acceleration for a 5 kg block.

What we did and why: - We isolated the block, drew only forces acting on it, and applied F = ma in the direction of motion. - Since there was no friction or vertical acceleration, we only needed the x-axis equation.

Example 2 – Medium (Inclined Plane with Friction)

Problem: A 10 kg block slides down a 30° incline with μ = 0.2. Find its acceleration.

Step 1: System & Frame

• Inertial frame (ground).
• No pseudo force.

Step 2: Free Body Diagram (FBD)

• Forces:
• mg = 10 × 9.8 = 98 N (downward).
• N (perpendicular to incline).
• f = μN (up the incline, opposing motion).
• Resolve mg into components:
• mg sin30° = 49 N (parallel to incline).
• mg cos30° = 84.87 N (perpendicular to incline).

Step 3: Resolve Forces

• x-axis (along incline): mg sin30° - f = ma.
• y-axis (perpendicular): N = mg cos30°.

Step 4: Apply F = ma

• y-axis: N = 84.87 N.
• Friction: f = μN = 0.2 × 84.87 = 16.97 N.
• x-axis: 49 - 16.97 = 10a → a = 3.203 m/s².

Step 5: Check

• Units: All forces in N, mass in kg → Correct.
• Reasonable: Acceleration less than g sin30° (4.9 m/s²) due to friction.

What we did and why: - We resolved mg into components because the motion is along the incline. - Friction opposes motion, so it’s subtracted from the driving force (mg sinθ). - Normal force ≠ mg on an incline—it’s mg cosθ.

Example 3 – Exam-Style (Pulley System with Pseudo Force)

Problem: A 4 kg block is connected to a 2 kg block via a massless string over a frictionless pulley. The 4 kg block is on a horizontal surface with μ = 0.3, and the 2 kg block hangs vertically. The entire system is in an elevator accelerating upward at 2 m/s². Find the tension in the string.

Step 1: System & Frame

• Non-inertial frame (elevator accelerating upward at 2 m/s²).
• Add pseudo force downward on both blocks.

Step 2: Free Body Diagrams (FBDs)

For 4 kg block (horizontal): - mg = 4 × 9.8 = 39.2 N (down). - N (up). - T (right, tension). - f = μN (left, friction). - Fₚₛₑᵤdₒ = 4 × 2 = 8 N (down, pseudo force).

For 2 kg block (vertical): - mg = 2 × 9.8 = 19.6 N (down). - T (up, tension). - Fₚₛₑᵤdₒ = 2 × 2 = 4 N (down, pseudo force).

Step 3: Resolve Forces

4 kg block (x-axis): - ΣFₓ = T - f = ma. - f = μN = 0.3 × (mg + Fₚₛₑᵤdₒ) = 0.3 × (39.2 + 8) = 14.16 N. - T - 14.16 = 4a (Equation 1).

2 kg block (y-axis): - ΣFᵧ = T - mg - Fₚₛₑᵤdₒ = ma. - T - 19.6 - 4 = 2a → T - 23.6 = 2a (Equation 2).

Step 4: Solve the System

• From Equation 1: T = 4a + 14.16.
• Substitute into Equation 2: (4a + 14.16) - 23.6 = 2a → 4a - 9.44 = 2a → 2a = 9.44 → a = 4.72 m/s².
• T = 4(4.72) + 14.16 = 33.04 N.

Step 5: Check

• Units: All forces in N, mass in kg → Correct.
• Reasonable: Tension less than 2 kg block’s weight (19.6 N)? No—because the elevator is accelerating upward, increasing effective weight.

What we did and why: - We treated the elevator as a non-inertial frame and added pseudo forces. - Friction depends on normal force, which increased due to the pseudo force. - Tension is the same throughout the string (massless string assumption).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is how you ace Newton’s laws in NEET:

1. Draw the FBD first—every force, every time. If the frame is accelerating (lift, car), add pseudo force opposite to acceleration.
2. Resolve forces into x and y—especially on inclines. mg sinθ drives motion, mg cosθ gives normal force.
3. Apply F = ma in each direction. If acceleration = 0, ΣF = 0. If accelerating, ΣF = ma.
4. Tension is uniform in massless strings. Spring force is F = -kx—watch the sign!
5. Check units and reasonableness—if acceleration is >10 m/s², you probably messed up.

Common traps? - Pseudo force in accelerating frames. - Normal force ≠ mg on inclines. - Friction opposes motion—don’t draw it wrong!

You’ve got this. Now go solve those problems like a pro."