Work Energy and Power — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Work Energy and Power" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Work Energy and Power" — 6 important questions with detailed answers for CBSE board ex…
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Key Questions Covered:
- Define work, energy, and power. Give SI units and one example of each.
- A force of 50 N is applied on a block at an angle of 30° above the horizontal…
- A ball of mass 0.5 kg is thrown vertically upward with an initial velocity of…
- + 3 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Define work, energy, and power. Give SI units and one exa… | ✓ Solved |
| A force of 50 N is applied on a block at an angle of 30° … | ✓ Solved |
| A ball of mass 0.5 kg is thrown vertically upward with an… | ✓ Solved |
Showing 3 of 6 questions
Q1: Define work, energy, and power. Give SI units and one example of each.
Definitions:
Work (W):
Work is defined as the product of force and displacement in the direction of the force.
W = F · d · cos θ
Where:
F = magnitude of force
d = magnitude of displacement
θ = angle between force and displacement
SI Unit: Joule (J) = N⋅m = kg⋅m²⋅s⁻²
Definition: One joule is the ...
Q2: A force of 50 N is applied on a block at an angle of 30° above the horizontal. The block moves 10 m horizontally. Calculate the work done by this force, the work done against gravity, and the net work if the block has mass 20 kg and g = 10 m/s².
Given:
Force F = 50 N
Angle with horizontal θ = 30°
Displacement s = 10 m (horizontal)
Mass m = 20 kg
g = 10 m/s²
Part 1: Work done by the applied force
The work done depends on the component of force in the direction of displacement.
Force component along displacement:
F_parallel = F cos θ = 50 ...
Q3: A ball of mass 0.5 kg is thrown vertically upward with an initial velocity of 20 m/s. Using energy conservation, find the maximum height reached. Also calculate the velocity when the ball returns to the point from which it was thrown. (g = 10 m/s²)
Given:
Mass m = 0.5 kg
Initial velocity u = 20 m/s (upward)
g = 10 m/s²
Part 1: Maximum height using energy conservation
At the point of throwing (initial position):
Height h₁ = 0 (reference level)
Velocity v₁ = 20 m/s
KE₁ = (1/2)mv₁² = (1/2)(0.5)(20)² = 0.25 × 400 = 100 J
PE₁ = 0 (at reference le...
Showing 3 of 6 questions. Visit the full page for complete solutions.