Test: Work Energy and Power - Class 10 MCQ

Formula: Work = Force × Displacement
Substitute: Work = 50 × 4
Calculation: Work = 200 J
Conclusion: All of the force acts along the displacement, so work done = 200 J.

Formula: Potential energy = m × g × h
Substitute: PE = 5 × 9.8 × 2
Calculation: PE = 5 × 19.6 = 98 J
Conclusion: Work done against gravity is stored as 98 J of gravitational potential energy.

Formula: Kinetic energy = 1/2 × m × v²
Substitute: KE = 1/2 × 2 × (3)²
Calculation: KE = 1 × 9 = 9 J
Conclusion: The moving mass has 9 J of kinetic energy.

Formula: Power = Work / Time
Substitute: P = 600 / 20
Calculation: P = 30 W
Conclusion: The man does work at a rate of 30 J every second.

Formula: Work = F × s × cosθ
Substitute: W = 10 × 2 × cos 60°
Calculation: cos 60° = 0.5 → W = 20 × 0.5 = 10 J
Conclusion: Only the component of force along displacement (10 N × 0.5) does work.

Formula: Work against gravity = m × g × h
Substitute: W = 10 × 9.8 × 5
Calculation: W = 10 × 49 = 490 J
Conclusion: 490 J of work must be done to lift the crate 5 m at constant speed.

Power = 2.0 kW, Time = 3 h → Energy (kWh) = Power × Time = 2.0 × 3 = 6.0 kWh.
Convert to joules: 1 kWh = 3.6 MJ → 6.0 kWh = 6 × 3.6 MJ = 21.6 MJ.
Conclusion: The heater uses 6 kWh, which equals 21.6 MJ of energy.

Work by brakes = change in kinetic energy = 1/2 m (v² - u²)
Substitute: 1/2 × 800 × (5² - 20²) = 400 × (25 - 400)
Calculation: 400 × (-375) = -150,000 J
Conclusion: Brakes do -150,000 J (negative indicates removing energy); magnitude = 150 kJ.

Work = m × g × h = 80 × 9.8 × 5 = 80 × 49 = 3920 J
Power = Work / Time = 3920 / 10 = 392 W
Conclusion: The lifting requires 392 W of power.

Work = F × s × cosθ = 20 × 3 × cos 30°
cos 30° = 0.866 → W = 60 × 0.866 = 51.96 J
Conclusion: Work done ≈ 51.96 J (use 51.96 J as precise value).

Find velocity: v = p / m = 20 / 5 = 4 m/s
KE = 1/2 × m × v² = 1/2 × 5 × 4² = 2.5 × 16 = 40 J
Conclusion: Kinetic energy = 40 J.

Elastic potential = 1/2 × k × x²
Substitute: = 1/2 × 200 × (0.10)² = 100 × 0.01 = 1.0 J
Conclusion: The stretched spring stores 1.0 J.

Work = m × g × h = 500 × 9.8 × 10 = 500 × 98 = 49,000 J
Power = Work / Time = 49,000 / 20 = 2450 W
Conclusion: Work = 49,000 J; power required = 2450 W.

Work by gravity when moving down = + m × g × h (force and displacement same direction)
W = 3 × 9.8 × 4 = 3 × 39.2 = 117.6 J
Conclusion: Gravity does +117.6 J of work on the mass while it falls 4 m.

hp = Power (W) / 746
Substitute: 2238 / 746 = 3.0 (exact)
Conclusion: 2238 W equals 3.0 horsepower.

Use energy conservation: mgh = 1/2 m v² → v = √(2 g h)
Substitute: v = sqrt(2 × 9.8 × 0.80) = sqrt(15.68) = 3.96 m/s (approx)
Conclusion: Speed at lowest point ≈ 3.96 m/s.

Work = F × s × cos 90° = F × s × 0 = 0 J
Perpendicular force does no work on that displacement.
Conclusion: Work done = 0 J.

Drop height so far = 10 - 6 = 4 m. Use v = √(2 g × drop)
v = sqrt(2 × 9.8 × 4) = sqrt(78.4) = 8.86 m/s (approx)
Conclusion: At 6 m height, speed ≈ 8.86 m/s downward.

Change in kinetic energy = 1/2 m (v_final² - v_initial²) = 1/2 × 1200 × (0 - 25²)
Calculation: 1/2 × 1200 = 600; 25² = 625 → Work = 600 × (-625) = -375,000 J
Conclusion: Friction does -375,000 J of work (removing kinetic energy).

Initial KE = 1/2 × m × v² = 1/2 × 2 × 6² = 1 × 36 = 36 J.
After +10 J work, final KE = 36 + 10 = 46 J.
Final speed v = √(2 × KE / m) = sqrt(2 × 46 / 2) = sqrt(46) ≈ 6.78 m/s.
Conclusion: New speed ≈ 6.78 m/s.