ISC Class 11 Physics Work-Energy Theorem Explained with Examples

Why the Work-Energy Theorem matters

Ever wondered why a roller‑coaster speeds up after the first drop? The answer lies in the work‑energy theorem – a handy shortcut that lets you connect forces and motion without solving messy equations.

💡 In Simple Words: The work‑energy theorem says that the total work a force does on an object changes that object's kinetic energy. In other words, push something and it speeds up; pull it and it slows down.

What is the Work‑Energy Theorem?

Work is the effort you apply over a distance. Imagine pushing a shopping cart: the force you exert multiplied by how far the cart moves in the direction of the force gives you the work done.

Kinetic energy (often abbreviated KE) is the energy an object has because it’s moving. It’s calculated as half the mass times the speed squared (½mv²). The first time you see these words, think of a moving car – the faster it goes, the more kinetic energy it stores.

The theorem itself states:

The net work done on a particle equals the change in its kinetic energy.

Mathematically, W_net = ΔK = K_final – K_initial. This simple relation works for any object as long as you consider all the forces that do work.

Deriving the theorem (quick peek)

Start with Newton’s second law: F = ma (force equals mass times acceleration). Multiply both sides by a tiny displacement dx:

F·dx = m a·dx

But a·dx can be rewritten as v·dv (because acceleration is the change in velocity over time). So we get:

F·dx = m v·dv

Now, F·dx is the infinitesimal work dW, and m v·dv is the infinitesimal change in kinetic energy dK. Integrating both sides from the start to the end of the motion gives the work‑energy theorem.

How to use the theorem – step by step

• Identify all forces that do work (push, pull, friction, gravity, etc.).
• Calculate the work each force does: W = F·s·cosθ where s is displacement and θ is the angle between force and displacement.
• Sum the works to get W_net.
• Find the change in kinetic energy using ΔK = W_net.
• If you know the initial speed, solve for the final speed, or vice‑versa.

Worked example 1 – A box on a frictionless ramp

A 5 kg box starts from rest at the bottom of a smooth 3‑m long ramp that makes a 30° angle with the horizontal. A constant horizontal push of 20 N is applied to the box. Find the speed of the box at the top of the ramp.

Solution

1. Break the push into components. The push is horizontal, but the box moves along the ramp. The angle between the push direction (horizontal) and the ramp direction is 30°. So the component of the push along the ramp is F_parallel = 20 cos30° ≈ 17.3 N.

2. Calculate work done by the push. Work = force × distance along the ramp: W_push = 17.3 N × 3 m ≈ 51.9 J.

3. Consider gravity. Gravity does negative work because it opposes the upward motion. The component of weight along the ramp is mg sin30° = 5×9.8×0.5 = 24.5 N downward. Work by gravity: W_g = -24.5 N × 3 m = -73.5 J.

4. Net work. W_net = W_push + W_g = 51.9 J – 73.5 J = -21.6 J. The negative sign means the box actually loses kinetic energy – the push isn’t strong enough to overcome gravity.

5. Find final speed. Initial kinetic energy is zero (starts from rest). So ΔK = W_net = -21.6 J = K_final – 0. Thus K_final = -21.6 J, which is impossible for kinetic energy. This tells us the box never reaches the top; it stops somewhere on the ramp. The example shows how the theorem can quickly reveal whether a motion is feasible.

Worked example 2 – Pulling a sled on snow

A child pulls a 10 kg sled with a constant force of 30 N making a 20° angle above the horizontal. The sled moves 5 m across level snow where kinetic friction exerts a force of 12 N opposite the motion. The sled starts from rest. Find its speed after 5 m.

Solution

1. Horizontal component of the pull: F_h = 30 cos20° ≈ 28.2 N.

2. Net horizontal force: Pull minus friction: F_net = 28.2 N – 12 N = 16.2 N.

3. Work done by the pull: W_pull = 30 N × 5 m × cos20° ≈ 141 J.

4. Work done by friction (negative): W_fric = -12 N × 5 m = -60 J.

5. Net work: W_net = 141 J – 60 J = 81 J.

6. Change in kinetic energy: ΔK = 81 J = K_final – 0. So K_final = 81 J.

7. Find final speed: K = ½mv² ⇒ 81 = ½×10×v² ⇒ v² = 16.2 ⇒ v ≈ 4.0 m/s.

Thus the sled is moving at about 4 m/s after the 5 m pull.

Quick comparison table

Key points to remember

• The theorem works for any object as long as you include all forces that do work.
• Positive work increases speed; negative work decreases it.
• If the net work is zero, the object's kinetic energy stays the same – it moves at constant speed.
• When friction or air resistance is present, treat them as forces that do negative work.

📝 Likely Exam Questions

1. State the work‑energy theorem and explain its significance.
   Answer: The work‑energy theorem states that the net work done on a particle equals the change in its kinetic energy (W_net = ΔK). It lets us connect forces directly to speed without solving detailed motion equations.
2. A 2 kg block is pushed across a frictionless table with a constant force of 10 N for 4 m. Find its final speed if it starts from rest.
   Answer: Work = 10 N × 4 m = 40 J. ΔK = 40 J = ½×2×v² ⇒ v² = 40 ⇒ v ≈ 6.32 m/s.
3. Explain why a car climbing a hill loses kinetic energy even though the engine is still delivering power.
   Answer: The engine’s force does positive work, but gravity does negative work larger than the engine’s work. Net work becomes negative, so kinetic energy decreases.
4. Calculate the work done by friction on a 5 kg box sliding 6 m with a kinetic friction force of 15 N opposite the motion.
   Answer: W_fric = -15 N × 6 m = -90 J.
5. Using the work‑energy theorem, determine the speed of a 0.5 kg ball thrown vertically upward with an initial speed of 10 m/s when it reaches its highest point.
   Answer: At the highest point, kinetic energy is zero. Net work by gravity = -mgh = -0.5×9.8×(10²/(2×9.8)) = -25 J, which equals the loss of initial kinetic energy (½mv² = 25 J). Hence speed is 0 m/s at the top.