Work Energy and Power Solved Examples (Class 11 Physics)
Work, energy, and power are interconnected concepts that describe how forces cause motion and transfer energy. These examples show how to calculate work do
TL;DR: Work, energy, and power are interconnected concepts that describe how forces cause motion and transfer energy. These examples show how to calculate wo…
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Work, energy, and power are interconnected concepts that describe how forces cause motion and transfer energy. These examples show how to calculate work do
Work Energy and Power — Solved Numerical Examples (Step by Step)
Example 1: A force of 20 N is applied to an object at an angle of 60° to the direction of motion. The object moves 5 m. Calculate the work done.
Solution: Given: Force F = 20 N, angle θ = 60°, displacement s = 5 m
Work done: W = F × s × cos θ
W = 20 × 5 × cos 60°
W = 20 × 5 × 0.5
W = 50 J
Example 2: A 2 kg object is lifted vertically upward by 10 m. Calculate the work done against gravity. (g = 10 m/s²)
Solution: Given: m = 2 kg, height h = 10 m, g = 10 m/s²
Work done against gravity = Force × displacement
= (Weight) × height
= mg × h
= 2 × 10 × 10
= 200 J
Example 3: A car of mass 1000 kg accelerates from rest to 20 m/s. Find the kinetic energy gained.
Solution: Given: m = 1000 kg, initial velocity u = 0, final velocity v = 20 m/s
Initial kinetic energy: KE_i = (1/2)mu² = (1/2) × 1000 × 0² = 0 J
Final kinetic energy: KE_f = (1/2)mv² = (1/2) × 1000 × 20² = 500 × 400 = 200000 J = 200 kJ
Kinetic energy gained = KE_f - KE_i = 200000 - 0 = 200000 J
Example 4: A ball of mass 0.5 kg is thrown upward with velocity 40 m/s. Find its potential energy at maximum height. (g = 10 m/s²)
Solution: Given: m = 0.5 kg, initial velocity u = 40 m/s, g = 10 m/s²
First, find maximum height using v² = u² - 2gh (at max height, v = 0)
0 = 40² - 2 × 10 × h
0 = 1600 - 20h
h = 1600/20 = 80 m
Potential energy at maximum height: PE = mgh
PE = 0.5 × 10 × 80 = 400 J
Example 5: An electric motor lifts a 50 kg load through a height of 20 m in 10 seconds. Calculate the power delivered by the motor. (g = 10 m/s²)
Solution: Given: m = 50 kg, h = 20 m, t = 10 s, g = 10 m/s²
Work done: W = mgh = 50 × 10 × 20 = 10000 J
Power: P = W/t = 10000/10 = 1000 W = 1 kW
Example 6: A vehicle of mass 1500 kg moves at constant velocity 15 m/s against a friction force of 3000 N. Find the power required to maintain this velocity.
Solution: Given: m = 1500 kg, v = 15 m/s (constant), friction force f = 3000 N
Since velocity is constant, the applied force equals friction force = 3000 N
Power: P = Force × velocity = f × v
P = 3000 × 15 = 45000 W = 45 kW
Tips
- Work done is zero if force is perpendicular to displacement (cos 90° = 0).
- When calculating total mechanical energy, always add kinetic and potential energy: E_total = KE + PE.
- Power can also be calculated as P = F × v when force is in the direction of motion.
- Use conservation of energy to solve problems: initial energy = final energy (in absence of non-conservative forces).
Frequently Asked Questions
Why is work a scalar quantity when force and displacement are vectors?
Work is defined as the dot product of force and displacement (W = F · s = Fs cos θ), which always produces a scalar value. This is because we only count the component of force in the direction of motion.
Can work be negative?
Yes, work can be negative. When the force applied is opposite to the direction of motion (angle > 90°), the work done is negative. For example, friction opposes motion, so it does negative work on the moving object.
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