Problem Types & Solved Examples
6 problem categories. For each: what the examiner tests, which concept to select, full solution, and the shortcut insight. This is exam preparation, not theory revision.
Every JEE problem on WEP falls into one of these 6 categories. Once you recognize the pattern, the approach becomes automatic. Pattern recognition beats memory every time.
Type 1: Direct Formula Application
Level: CBSE / NEET | W = Fd cos θ, KE = ½mv², P = W/t
A force of 20 N acts on a box and moves it 5 m at 60° to the force. Find work done.
What examiner tests: Can you correctly identify θ?
Concept: W = Fd cos θ
- 1W = 20 × 5 × cos 60° = 100 × 0.5 = 50 J
Shortcut: cos 0° = 1, cos 60° = 0.5, cos 90° = 0, cos 120° = −0.5. Memorize these four values — they cover 90% of NEET problems.
A pump lifts 200 kg of water per minute to a height of 10 m. Find power. (g = 10 m/s²)
What examiner tests: Can you convert minutes to seconds?
- 1
Work done in lifting water:
W = mgh = 200 × 10 × 10 = 20,000 J - 2P = W/t = 20,000/60 ≈ 333 W
Mistake: Using t = 1 minute = 1 s (wrong). Always convert to SI: 1 minute = 60 seconds.
Type 2: Conceptual (Zero Work, Sign Analysis)
Level: NEET / JEE Main | Requires understanding over calculation
A satellite moves in a circular orbit. Which of the following is correct about the work done by gravity?
What examiner tests: Does gravity do work in circular motion?
In circular orbit: gravity is centripetal force — always pointing toward center (radius direction). Velocity is always tangential (perpendicular to radius). Therefore, angle between gravity and velocity = 90° always. W = Fv cos 90° = 0.
Answer: Gravity does ZERO work on satellite in circular orbit.
A person carries a heavy load and walks on a horizontal floor. Does normal force or gravity do work?
What examiner tests: Understanding of when force does work
Displacement: Horizontal. Normal force: Vertical (upward). Gravity: Vertical (downward). Both forces are ⊥ to displacement. Therefore both do zero work. Note: The person doing the work uses internal (chemical) energy, not mechanical work input.
Answer: Both gravity and normal force do zero work.
This appears in NEET almost every 3 years. The person feels tired because muscles use chemical energy, not mechanical work.
Type 3: Multi-Step Energy Conservation
Level: JEE Main / Advanced | 4–6 step problems
A block (m = 2 kg) is released from the top of a smooth incline (height h = 5 m). At the bottom, it travels on a rough horizontal surface (μ = 0.2) and then compresses a spring (k = 1000 N/m). Find the maximum compression of the spring. (g = 10 m/s²)
What examiner tests: Can you track energy through 3 different phases?
Concept: Energy conservation across all phases, including friction loss on horizontal surface only.
- 1
PE at top converted to KE at bottom (smooth incline — no friction):
½mv²_bottom = mgh = 2×10×5 = 100 Jv_bottom = √100 = 10 m/s - 2
At max compression x: Block is at rest. KE = 0. Spring PE = ½kx²
But friction acts over distance d (horizontal surface). Let's call total horizontal distance traveled = D (rough surface) + x (spring compression):
Assuming rough surface of length D = 3 m (say) - 3
Energy equation (from top to max compression):
mgh = ½kx² + μmg(D + x)100 = 500x² + 0.2×2×10×(3+x)100 = 500x² + 4(3+x)100 = 500x² + 4x + 12500x² + 4x − 88 = 0 - 4
Using quadratic formula: x = (−4 + √(16 + 4×500×88))/(2×500)
x ≈ 0.41 m
Always define a clear initial and final state. Initial = block at rest at top. Final = spring maximally compressed (block at rest). All energy lost = friction work. This 3-phase structure is the standard JEE approach.
Don't break it into 2 separate problems. Always apply one master energy equation from initial to final state. This avoids the round-off errors that accumulate if you calculate intermediate velocities.
Common mistake: Forgetting friction during spring compression. Friction acts on the full path including the compression distance x.
Type 4: Graph-Based Problems
Level: JEE Main / Advanced | F-x, KE-x, PE-x, P-t graphs
F-x Graph Problems
Work = area under F-x graph between two positions. This is the most tested graph type.
W_total = Area 1 (positive) + Area 2 (negative)
If F-x is a triangle: Area = ½ × base × height. If trapezoid: Area = ½(F₁+F₂)×d. If curve: use integration. In JEE Main, graphs are almost always straight lines — use geometry.
KE-x Graph: Reading Velocity
If a KE-x graph is given, slope = F = ma (from W-E theorem). You can find force, acceleration, and work from such a graph.
If KE vs x graph is a straight line through origin: KE = cx → ½mv² = cx → v ∝ √x. The force F = dKE/dx = c (constant). This is a uniform force scenario.
Power-Time Graph
Work done = area under P-t graph. If P-t is a rectangle: W = P × t. If P-t is a triangle: W = ½ × base × height.
JEE Main pattern (appears almost every year): A force varies as F = kx. Find work done from x = 0 to x = d. This is an F-x graph problem. W = area of triangle = ½ × kd × d = ½kd². This is the origin of spring PE formula!
Type 5: Assertion & Reason Questions
Level: NEET / JEE Main | Tests conceptual clarity
Assertion (A):
Work done by friction can be positive.
Reason (R):
Friction always opposes relative motion between surfaces.
Assertion: TRUE. Static friction can do positive work. Example: A book kept on a truck moving forward — static friction on book acts forward (same direction as motion) → positive work.
Reason: TRUE. Friction does oppose relative motion (or tendency of relative motion). But this does NOT mean it always does negative work on an individual body — only on the slipping point.
Answer: Both A and R are true, but R is NOT the correct explanation of A.
Assertion (A):
The work done by centripetal force is always zero.
Reason (R):
The centripetal force acts along the radius, perpendicular to velocity.
Assertion: TRUE. Centripetal force ⊥ velocity → W = 0.
Reason: TRUE. Centripetal force acts radially inward. Velocity is tangential. These are always perpendicular in circular motion.
R correctly explains why W = 0 (because cos 90° = 0).
Answer: Both A and R are true, and R IS the correct explanation of A.
Type 6: Case-Based / Paragraph Problems
Level: CBSE Boards / JEE Advanced | Multi-question from one scenario
Paragraph / Case
A block of mass 4 kg starts from rest and slides down a smooth curved surface of height 2 m. It then slides on a horizontal rough surface (μ = 0.25, length = 4 m) and finally strikes a spring (k = 500 N/m) attached to a wall. (g = 10 m/s²)
Q1: Speed at bottom of incline
- 1
Smooth incline → No friction. Energy conservation:
½mv² = mgh = 4×10×2 = 80 Jv = √40 = 6.32 m/s
Q2: Speed after crossing rough surface
- 1
Friction work = μmgd = 0.25×4×10×4 = 40 J
KE_2 = 80 − 40 = 40 J → v₂ = √20 ≈ 4.47 m/s
Q3: Maximum spring compression
- 1
At max compression: KE = 0. Spring PE = ½kx²
40 = ½ × 500 × x²x = √(80/500) = √0.16 = 0.4 m
Q4: Block returns — final position?
Block bounces back. Spring gives 40 J. Friction acts again for 4 m. Final KE = 40 − 40 = 0 J. Block just stops at the start of rough surface. This is a CBSE board question type — tests whether students apply friction twice.
Case-based strategy: Always draw the scenario with energy values at each point. Fill in: PE → KE → friction loss → spring PE → back. This gives you all answers without solving separately.