Vidhyapath← NEET Hub
🧪 Mocks
PHY 01Physics4–5 Q/year★★★★★ Highest Priority
Laws of Motion & Work-Energy
NCERT Class 11 Ch. 4–6. Newton's laws, friction, work-energy theorem — 4–5 Q per year, most scoring.
4–5
Questions/Year
16–20
Marks at Stake
Laws of Motion & Work-Energy
Newton's laws and work-energy theorem form the backbone of NEET Physics. Friction and connected bodies are the most tested application areas.
Newton's Laws

First law (Inertia): An object continues in its state of rest or uniform motion unless acted upon by a net external force. Inertia is proportional to mass.

Second law: F = ma (F_net = rate of change of momentum = dp/dt). Unit: Newton = kg·m/s².

Third law: For every action, there is an equal and opposite reaction. Forces in action-reaction pairs act on DIFFERENT bodies.

Free Body Diagram (FBD): Draw all forces on the body of interest. Normal force (N) perpendicular to surface. Weight (mg) downward. Tension (T) along string. Friction (f) opposing motion.

Friction

Static friction: f_s ≤ μ_s N. Adjusts to oppose tendency of motion. Maximum static friction = μ_s N (limiting friction).

Kinetic friction: f_k = μ_k N. Constant once sliding. μ_k < μ_s (easier to keep moving than to start).

Inclined plane: Body on incline angle θ. N = mg cos θ. Net force down incline = mg sin θ - f = mg sin θ - μ_k mg cos θ = mg(sin θ - μ_k cos θ). Angle of repose: tan θ = μ_s (body just begins to slide).

Connected bodies: Block A (m₁) and B (m₂) on surface connected by string. Pulling force F on B. Tension T = m₁F/(m₁+m₂). Acceleration a = F/(m₁+m₂).

Work, Energy & Power

Work: W = F·d·cos θ. Positive work: F has component in direction of displacement. Zero work: perpendicular force (centripetal force). Negative work: F opposes displacement (friction, braking).

Kinetic energy: KE = ½mv². Work-energy theorem: W_net = ΔKE = ½mv² - ½mu².

Potential energy: Gravitational PE = mgh. Spring PE = ½kx². Conservative forces: work done independent of path; work done in closed path = 0.

Conservation of energy: KE + PE = constant (no non-conservative forces). Power P = W/t = Fv.

Collisions

Elastic collision: Both KE and momentum conserved. For equal masses: velocities exchange. For m₁ >> m₂: heavy body barely slows, light body moves at 2× heavy body's speed.

Inelastic collision: Momentum conserved, KE NOT conserved. Coefficient of restitution e = 1 (elastic), e = 0 (perfectly inelastic — bodies stick together).

Perfectly inelastic: v_common = (m₁v₁ + m₂v₂)/(m₁+m₂). Maximum KE loss occurs in perfectly inelastic collision.

Laws of Motion Formulas
Friction Equations
Static: f_s ≤ μ_s N
Kinetic: f_k = μ_k N
Always: μ_k < μ_s
Angle of repose: tan θ = μ_s
Incline: N = mg cos θ
Friction acts parallel to surface, opposing relative motion or tendency of motion
Work-Energy
W = F·d·cos θ
KE = ½mv²
PE (gravity) = mgh
PE (spring) = ½kx²
W_net = ΔKE
Power P = W/t = Fv
Work done by gravity = -ΔPE = mgh (taking down as positive). Path independent.
Collision Formulas
Elastic: momentum + KE conserved
v₁' = (m₁-m₂)v₁/(m₁+m₂) [if v₂=0]
v₂' = 2m₁v₁/(m₁+m₂)
Inelastic: v = (m₁v₁+m₂v₂)/(m₁+m₂)
e = 1 (elastic), e = 0 (perfectly inelastic)
In perfectly inelastic collision, loss in KE = ½[m₁m₂/(m₁+m₂)](v₁-v₂)²
Atwood Machine
Two masses m₁, m₂ over pulley (m₁>m₂):
a = (m₁-m₂)g/(m₁+m₂)
T = 2m₁m₂g/(m₁+m₂)
Net force = (m₁-m₂)g
Total mass = m₁+m₂
If m₁ = m₂: a = 0, T = m₁g = m₂g (static equilibrium)
Worked Examples
EasyA 5 kg block is pushed along a horizontal surface with force 30N at angle 0°. μ_k = 0.2. Find acceleration. (g=10)
N = mg = 5×10 = 50N. f_k = μ_k N = 0.2×50 = 10N. Net force = 30 - 10 = 20N. a = F_net/m = 20/5 = 4 m/s².
Answer: a = 4 m/s²
MediumIn Atwood machine: m₁ = 6 kg, m₂ = 4 kg. Find acceleration and tension. (g=10 m/s²)
a = (m₁-m₂)g/(m₁+m₂) = (6-4)×10/(6+4) = 20/10 = 2 m/s². T = 2m₁m₂g/(m₁+m₂) = 2×6×4×10/10 = 48 N.
Answer: a = 2 m/s², T = 48 N
Mistake DNA
❌ Applying net force = ma to individual components without FBD
Students apply ma to the "system" and get the acceleration but then use the same total force to find tension — wrong. The tension on each block depends only on the net force acting on THAT block, found by drawing separate FBDs.
Fix: Draw separate FBD for each object. Apply F=ma to each separately. Tension appears in both FBDs.
❌ Saying static friction always equals μN
Static friction adjusts to MATCH the applied force (up to maximum μ_s N). If the applied force is less than the maximum, f_s < μ_s N. Only at the point of slipping does f_s = μ_s N.
Fix: f_s ≤ μ_s N. The ≤ is critical. Static friction is reactive, not always maximum.
Chapter Intelligence
PYQ Frequency
FBD/friction problems: 1–2 Q/year
Work-energy: 1–2 Q/year
Collision: 1 Q/year
Atwood/connected bodies: 1 Q/year
2026 Prediction
High: Work-energy theorem application
Expected: Friction on inclined plane
Watch: Elastic collision velocity calculation
Practice in NEET Full Mocks
Start NEET Mock 1 — Free →
Share

💡 Suggestions & Feedback

We read every message

To keep feedback accountable, we verify your email before accepting messages.

Already a student? Log in to skip this step.