Work, Energy and Power — Flashcards
NIOS · Class 12 · Physics
39 flashcards for Work, Energy and Power (NIOS Class 12 Physics) to test yourself on key terms and facts. Part of the NIOS Class 12 Physics syllabus.
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State the definition and formula for Work done by a constant force. What is its SI unit?
Answer
Work done by a constant force F acting at angle θ with displacement d is: W = F · d · cos θ In vector form: W = F⃗ · d⃗ Key Points: • Work is a SCALAR quantity • If θ = 0°, W = Fd (maximum positive…
A force of 6 N is applied on an object at 60° with the horizontal. The object moves 2 m horizontally. Calculate the work done.
Answer
Given: F = 6 N, θ = 60°, d = 2 m Find: Work done W Formula: W = F · d · cos θ Solution: W = 6 × 2 × cos 60° W = 6 × 2 × 0.5 W = 6 J Answer: W = 6 J Note: Only the horizontal component (F cos60° =…
Why is no work done when you push a wall with all your strength but it does not move?
Answer
Because Work = F × d × cos θ, and displacement d = 0. W = F × 0 × cos θ = 0 J Even if force applied is very large, ZERO displacement means ZERO work done in the scientific sense. Real-world example…
A force F⃗ = (2î + 3ĵ) N produces a displacement d⃗ = (–î + 2ĵ) m. Calculate the work done.
Answer
Given: F⃗ = (2î + 3ĵ) N d⃗ = (–î + 2ĵ) m Formula: W = F⃗ · d⃗ Solution (Dot Product): W = (2)(–1) + (3)(2) W = –2 + 6 W = 4 J Answer: W = 4 J Note: Even though one component contributes negative w…
When is work done by gravity POSITIVE and when is it NEGATIVE? Explain with examples.
Answer
Work done by Gravity = mg × displacement × cos θ CASE 1 – Object moving DOWNWARD: • Force (mg) ↓ and displacement ↓ → θ = 0° • W_gravity = +mgh (POSITIVE) • Example: A stone falling from a building …
State Hooke's Law for a spring. Write the formula for work done in compressing or stretching a spring by distance x.
Answer
Hooke's Law: The restoring force in a spring is proportional to the displacement from equilibrium: F_s = –kx Where: • k = spring constant (N/m) • x = compression or extension (m) • Negative sign sho…
A mass of 2 kg is attached to a spring of force constant k = 100 N/m. Calculate the work done by an external force in stretching the spring by 10 cm.
Answer
Given: m = 2 kg, k = 100 N/m, x = 10 cm = 0.1 m Find: Work done W Formula: W = ½ k x² Solution: W = ½ × 100 × (0.1)² W = ½ × 100 × 0.01 W = 50 × 0.01 W = 0.5 J Answer: W = 0.5 J Note: Work done b…
Define Power. Write its formula, SI unit, and dimensional formula. How is kilowatt-hour (kWh) related to joules?
Answer
Power is the RATE of doing work. Average Power: P = W / t Instantaneous Power: P = dW/dt SI Unit: Watt (W) = Joule/second = J s⁻¹ Dimensional Formula: [ML²T⁻³] Other Units: • 1 kW = 10³ W • 1 MW …
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