By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Mastering Work, Energy & Power unlocks 8-10 marks in NEET Physics—enough to push you into the top 1%. From calculating how much force a motor needs to lift an elevator to predicting the speed of a rollercoaster at the bottom of a drop, this topic bridges theory and real-world engineering. If you can solve these problems, you’re not just passing—you’re dominating.
Before diving in, ensure you understand: 1. Newton’s Second Law (F = ma) – How forces cause acceleration. 2. Kinetic Energy (KE = ½mv²) & Potential Energy (PE = mgh) – The two most common forms of mechanical energy. 3. Basic Calculus (for Power = dW/dt) – Only differentiation of work with respect to time is needed.
If any of these are shaky, pause and review them first.
Problem: A 2 kg block slides from rest down a frictionless incline of height 5 m. What is its speed at the bottom?
Solution: 1. Identify forces: Only gravity (conservative). 2. Approach: Conservation of mechanical energy (KE + PE = constant). 3. Known: - m = 2 kg - h = 5 m - g = 9.8 m/s² - Initial speed (u) = 0 → KE_initial = 0 4. At top: PE = mgh = 2 × 9.8 × 5 = 98 J 5. At bottom: PE = 0, KE = ½mv² 6. Apply conservation: KE_initial + PE_initial = KE_final + PE_final 0 + 98 = ½ × 2 × v² + 0 98 = v² → v = √98 ≈ 9.9 m/s
What we did and why: - Used conservation of energy because only conservative forces act. - Ignored friction (given as frictionless), so no energy loss.
Problem: A 1000 kg car accelerates from 10 m/s to 20 m/s over 50 m. If the engine exerts a constant force of 2000 N, what is the work done by friction?
Solution: 1. Identify forces: Engine force (non-conservative), friction (non-conservative). 2. Approach: Work-Energy Theorem (W_net = ΔKE). 3. Known: - m = 1000 kg - u = 10 m/s, v = 20 m/s - s = 50 m - F_engine = 2000 N 4. Calculate ΔKE: ΔKE = ½mv² – ½mu² = ½ × 1000 × (20² – 10²) = 500 × 300 = 150,000 J 5. Work done by engine: W_engine = F_engine × s = 2000 × 50 = 100,000 J 6. Net work = ΔKE: W_net = W_engine + W_friction = 150,000 100,000 + W_friction = 150,000 → W_friction = 50,000 J (Since friction opposes motion, W_friction = -50,000 J)
What we did and why: - Used Work-Energy Theorem because non-conservative forces (engine, friction) are involved. - Friction does negative work (opposes motion).
Problem: A motor lifts a 500 kg elevator at a constant speed of 2 m/s. If the motor’s efficiency is 80%, what is the input power?
Solution: 1. Identify forces: Gravity (conservative), motor force (non-conservative). 2. Approach: Power = F · v, then account for efficiency. 3. Known: - m = 500 kg - v = 2 m/s (constant speed → a = 0 → F_net = 0) - g = 9.8 m/s² - η = 80% = 0.8 4. Force needed to lift at constant speed: F = mg = 500 × 9.8 = 4900 N 5. Output power (useful power): P_out = F · v = 4900 × 2 = 9800 W 6. Input power (since η = P_out / P_in): P_in = P_out / η = 9800 / 0.8 = 12,250 W
What we did and why: - At constant speed, net force = 0 → motor force = weight. - Used P = F · v for instantaneous power. - Efficiency relates input power to output power.
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