Work and Energy Solved Examples (Class 9 Physics)
Work is done when a force moves an object. Energy is the capacity to do work. Understanding kinetic energy, potential energy, and conservation of energy is
TL;DR: Work is done when a force moves an object. Energy is the capacity to do work. Understanding kinetic energy, potential energy, and conservation of ener…
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Work is done when a force moves an object. Energy is the capacity to do work. Understanding kinetic energy, potential energy, and conservation of energy is
Work and Energy — Solved Numerical Examples (Step by Step)
Example 1: A force of 50 N is applied to move an object 20 m in the direction of the force. Calculate the work done.
Solution: Given:
Force F = 50 N
Displacement s = 20 m
Angle θ = 0 degrees (force in direction of displacement)
Using the formula: Work W = F × s × cos(θ)
W = 50 × 20 × cos(0)
W = 50 × 20 × 1
W = 1000 J
Example 2: Calculate the kinetic energy of a 5 kg object moving at 10 m/s.
Solution: Given:
Mass m = 5 kg
Velocity v = 10 m/s
Using the formula: Kinetic energy KE = (1/2)mv^2
KE = (1/2) × 5 × (10)^2
KE = (1/2) × 5 × 100
KE = 2.5 × 100
KE = 250 J
Example 3: A 2 kg object is lifted to a height of 5 m above the ground. Calculate its gravitational potential energy. (g = 10 m/s^2)
Solution: Given:
Mass m = 2 kg
Height h = 5 m
Acceleration due to gravity g = 10 m/s^2
Using the formula: Potential energy PE = mgh
PE = 2 × 10 × 5
PE = 100 J
Example 4: A ball is dropped from a height of 20 m. What is its velocity when it reaches the ground? (g = 10 m/s^2)
Solution: Given:
Height h = 20 m
Initial velocity u = 0 m/s (dropped from rest)
g = 10 m/s^2
Using conservation of energy: PE at top = KE at bottom
mgh = (1/2)mv^2
gh = (1/2)v^2
10 × 20 = (1/2)v^2
200 = (1/2)v^2
v^2 = 400
v = 20 m/s
Example 5: Calculate the power of a crane that lifts a 100 kg load to a height of 10 m in 5 seconds. (g = 10 m/s^2)
Solution: Given:
Mass m = 100 kg
Height h = 10 m
Time t = 5 seconds
g = 10 m/s^2
Work done W = mgh = 100 × 10 × 10 = 10000 J
Using the formula: Power P = W / t
P = 10000 / 5
P = 2000 W = 2 kW
Example 6: A 3 kg object at rest is pushed by a force of 60 N for 4 m. Find its final kinetic energy assuming no friction.
Solution: Given:
Mass m = 3 kg
Force F = 60 N
Displacement s = 4 m
Initial velocity u = 0 (at rest)
Work done by force W = F × s = 60 × 4 = 240 J
By work-energy theorem: Work done = Change in kinetic energy
240 = KEfinal - KEinitial
240 = KEfinal - 0
KEfinal = 240 J
Example 7: A car of mass 1000 kg is moving at 20 m/s. Its brakes apply a force that brings it to rest in 10 seconds. Calculate the average power dissipated.
Solution: Given:
Mass m = 1000 kg
Initial velocity u = 20 m/s
Final velocity v = 0 m/s
Time t = 10 seconds
Initial kinetic energy KE1 = (1/2) × 1000 × (20)^2 = 500 × 400 = 200000 J
Final kinetic energy KE2 = 0 J
Energy dissipated = 200000 J
Average power P = Energy / Time = 200000 / 10 = 20000 W = 20 kW
Tips
- Work is a scalar quantity. It is positive when force and displacement are in the same direction, zero when perpendicular, and negative when opposite. Use W = F × s × cos(θ) when force is at an angle.
- Energy cannot be created or destroyed, only converted from one form to another. In any system without external friction, total mechanical energy (KE + PE) remains constant.
- Power is the rate of doing work. P = W/t measures how fast work is done. A larger power means more work in less time.
Frequently Asked Questions
What is the work-energy theorem?
The work-energy theorem states that the total work done by all forces on an object equals the change in its kinetic energy. Mathematically, W_total = KE_final - KE_initial. This is useful for finding the final velocity or work done without knowing the path taken.
How are energy and power related?
Energy is the capacity to do work (measured in Joules). Power is the rate at which energy is transferred or work is done (measured in Watts). Power = Energy / Time. For example, a 100 W bulb uses 100 Joules of electrical energy every second.
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