Work Energy Theorem – Class 11 + JEE Main | Complete Concept, Formula & Solved Question

Work Energy Theorem – Class 11 + JEE Main

Work, Energy and Power is one of the most important chapters in Class 11 Physics and is highly useful for JEE Main. Many questions that look lengthy can actually be solved in just a few lines using the Work-Energy Theorem.

The central idea is very simple:

NET WORK DONE = CHANGE IN KINETIC ENERGY

Mathematically:

W_net = ΔK

This theorem allows us to connect force, displacement and velocity without necessarily calculating acceleration or time.

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1. Work — Basic Idea

When a force acts on an object and the object undergoes displacement, the force may do work on the object.

For a constant force:

W = Fs cosθ

where:

• F = magnitude of force
• s = displacement
• θ = angle between force and displacement

The sign of work depends on the angle between the force and displacement.

Special Cases

Angle	Work	Meaning
θ = 0°	W = +Fs	Force and displacement are in the same direction
θ = 90°	W = 0	Force is perpendicular to displacement
θ = 180°	W = −Fs	Force and displacement are opposite

Therefore, work can be positive, negative or zero.

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2. Kinetic Energy ⭐

Kinetic energy is the energy possessed by an object because of its motion.

K = ½mv²

where:

• m = mass of the object
• v = speed of the object

The SI unit of kinetic energy is the Joule (J).

If the velocity changes from u to v, then the change in kinetic energy is:

ΔK = ½mv² − ½mu²

Therefore:

ΔK = ½m(v² − u²)

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3. Work-Energy Theorem 🔥

The Work-Energy Theorem states that the net work done by all forces acting on an object is equal to the change in its kinetic energy.

W_net = ΔK

Therefore:

W_net = ½mv² − ½mu²

or:

W_net = ½m(v² − u²)

This is the most important formula of this topic.

What does it tell us?

• Positive net work → kinetic energy increases.
• Negative net work → kinetic energy decreases.
• Zero net work → kinetic energy remains constant.

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4. Derivation of Work-Energy Theorem ⭐

For motion with constant acceleration, we know:

v² − u² = 2as

Multiply both sides by m/2:

½m(v² − u²) = mas

From Newton's second law:

F_net = ma

Therefore:

mas = F_nets

So:

½m(v² − u²) = F_nets

But:

F_nets = W_net

Hence:

W_net = ½m(v² − u²) = ΔK

This is the Work-Energy Theorem.

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5. The Most Important Concept ⚠️

The theorem does not say that the work done by any one particular force is always equal to the change in kinetic energy.

It uses the net work done by all forces.

If several forces act on an object:

W_net = W₁ + W₂ + W₃ + ...

Then:

W_net = ΔK

This distinction is extremely important in JEE Main questions involving friction, gravity, normal reaction and applied forces together.

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6. Example — Horizontal Block

A block of mass 2 kg accelerates from 3 m/s to 7 m/s. Find the net work done.

Given:

• m = 2 kg
• u = 3 m/s
• v = 7 m/s

Using:

W_net = ½m(v² − u²)

W_net = ½(2)(49 − 9)

W_net = 40 J

✅ W_net = 40 J

Notice that we did not need to calculate acceleration or time.

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7. Work Done by Individual Forces

Consider a block moving horizontally under different forces.

Applied Force

If the applied force acts in the direction of displacement:

W_F = +Fs

So the applied force does positive work.

Friction

For ordinary sliding motion, friction acts opposite to displacement:

W_f = −fs

Therefore, kinetic friction usually does negative work on the sliding block.

Normal Reaction

For horizontal displacement, the normal reaction is perpendicular to the displacement:

W_N = 0

Gravity

For horizontal displacement, gravity is also perpendicular to the displacement:

W_g = 0

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8. Friction + Work-Energy Theorem 🔥

Suppose a block moves on a rough horizontal surface.

Let:

• Applied force = F
• Friction = f
• Displacement = s

The net work is:

W_net = Fs − fs

Therefore:

W_net = (F − f)s

Using the Work-Energy Theorem:

(F − f)s = ½m(v² − u²)

This equation is very useful for questions involving a block being pulled or pushed along a rough surface.

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9. Work-Energy Theorem on an Inclined Plane

Consider a block moving down an inclined plane of angle θ.

The component of gravitational force along the plane is:

mg sinθ

If friction acts upward along the plane, its magnitude is f.

Therefore, the net force along the plane is:

mg sinθ − f

For displacement s along the plane:

W_net = (mg sinθ − f)s

Hence:

(mg sinθ − f)s = ΔK

The Work-Energy approach is often faster than using equations of motion when only initial and final speeds are required.

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10. Work Done by Gravity ⭐

Gravity is a conservative force. Its work depends only on the initial and final heights.

The work done by gravity is:

W_g = mg(h_i − h_f)

where:

• h_i = initial height
• h_f = final height

Body Moving Down

If the body falls through height h:

W_g = +mgh

Body Moving Up

If the body moves upward through height h:

W_g = −mgh

JEE Shortcut:

Down → +mgh

Up → −mgh

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11. Work Done by Spring

For an ideal spring, the restoring force is given by:

F = −kx

The negative sign indicates that the spring force acts opposite to the displacement from its equilibrium position.

The work done by the spring when the position changes from x₁ to x₂ is:

W_s = ½k(x₁² − x₂²)

The elastic potential energy stored in the spring is:

U_s = ½kx²

So, when the spring is stretched or compressed, it can store mechanical energy and later transfer that energy to the object.

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12. Conservative Force ⭐

A force is called conservative when the work done by it is independent of the path followed between the initial and final positions.

In simple words:

Same initial point + same final point → same work

Important examples include:

• Gravitational force
• Spring force
• Electrostatic force

Closed Path

For a conservative force, the total work done over a closed path is:

W_closed = 0

This is an important property that frequently appears in conceptual questions.

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13. Non-Conservative Force

For a non-conservative force, the work done generally depends on the path followed.

A standard example is:

Friction

For example, if an object slides a longer distance because of a different path, the magnitude of frictional work generally changes because friction acts over the distance travelled.

Mechanical energy is generally dissipated as thermal energy when friction acts.

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14. What Does the Sign of Net Work Tell Us?

Case 1: W_net > 0

Positive net work means:

ΔK > 0

Therefore kinetic energy increases and the object's speed increases.

Case 2: W_net < 0

Negative net work means:

ΔK < 0

Therefore kinetic energy decreases and the object's speed decreases.

Case 3: W_net = 0

Zero net work means:

ΔK = 0

Therefore kinetic energy remains constant and the speed remains constant.

Important: Constant speed does not necessarily mean constant velocity. The direction of motion can still change.

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15. Work-Energy and F-x Graph 🔥

For a variable force, work cannot generally be calculated using only Fs.

Instead:

W = ∫F dx

Graphically, the work done is represented by the area under the F-x graph.

Therefore:

Area under F-x graph = Work

For the net force:

Area under F_net-x graph = ΔK

This is a very important JEE Main graphical concept.

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16. Variable Force

When force changes with position, the work done is:

W = ∫ F dx

For example, if the force varies with position, we can calculate the work by finding the area under the corresponding force-displacement curve.

Remember:

Constant Force → W = Fs cosθ

Variable Force → W = ∫F dx

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17. Power ⭐

Power tells us the rate at which work is done.

Average power is:

P = W/t

For instantaneous power:

P = F · v

If force and velocity make an angle θ:

P = Fv cosθ

The SI unit of power is the Watt (W).

Thus, power tells us how quickly work is being performed, while work tells us the total energy transferred through a force.

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18. JEE Shortcut 🔥

One of the biggest advantages of the Work-Energy Theorem is that it can eliminate unnecessary calculations.

If the question gives only initial and final speeds, directly use:

W_net = ½m(v² − u²)

Do not calculate acceleration unless the question actually requires it.

If height changes are involved:

W_g = −ΔU

For a spring:

U_s = ½kx²

For a variable force:

W = ∫F dx

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19. JEE Main-Level Solved Question 🔥

Question: A 2 kg block is moving with a speed of 4 m/s on a horizontal surface. A constant horizontal force does 50 J of work on the block, while friction does 18 J of work. Find the final speed.

Step 1: Calculate Net Work

The applied force does positive work:

W_F = +50 J

Friction does negative work:

W_f = −18 J

Therefore:

W_net = 50 − 18

W_net = 32 J

Step 2: Apply Work-Energy Theorem

W_net = ½m(v² − u²)

Substitute:

32 = ½(2)(v² − 16)

32 = v² − 16

v² = 48

Therefore:

v = 4√3 m/s

✅ Final Answer = 4√3 m/s

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20. Common JEE Traps ⚠️

Trap 1: Using Work Done by One Force

Wrong idea:

Work done by one force = ΔK

Correct:

Net work done by all forces = ΔK

Trap 2: Assuming Normal Force Always Does Work

Work depends on the angle between force and displacement.

If the normal force is perpendicular to displacement:

W_N = 0

But one should not blindly say that a normal force always does zero work in every possible situation.

Trap 3: Assuming Friction Always Does Negative Work

For ordinary sliding motion, kinetic friction on the sliding object generally does negative work. However, friction is not universally guaranteed to do negative work on every individual body; its work depends on the actual motion and point of application.

Trap 4: Zero Work Means Zero Force

Incorrect.

A force can be non-zero while doing zero work if it is perpendicular to displacement.

Example:

Uniform circular motion → centripetal force is perpendicular to instantaneous displacement → instantaneous work is zero.

Trap 5: Work is Always Positive

Incorrect.

Work can be:

• Positive
• Negative
• Zero

Trap 6: Confusing Kinetic Energy with Potential Energy

The Work-Energy Theorem directly gives:

Change in kinetic energy

not directly the change in potential energy.

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21. JEE Main Quick-Solving Strategy

1. List all forces acting on the object.
2. Determine the work done by each force.
3. Add all individual works to obtain net work.
4. Use W_net = ΔK.
5. If only initial and final speeds are given, directly use ½m(v² − u²).
6. If height changes, carefully calculate the work done by gravity.
7. If a spring is present, use U = ½kx².
8. If force varies with position, use the F-x graph or integration.

This method is especially useful when Newton's laws would require multiple equations but the question only asks for speed or energy.

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22. Work-Energy Formula Sheet 📌

Concept	Formula
Work by constant force	W = Fs cosθ
Kinetic energy	K = ½mv²
Change in kinetic energy	ΔK = ½m(v² − u²)
Work-Energy Theorem	Wnet = ΔK
Work by gravity	Wg = mg(hi − hf)
Spring force	F = −kx
Spring potential energy	Us = ½kx²
Variable-force work	W = ∫F dx
F-x graph	Area = Work
Average power	P = W/t
Instantaneous power	P = F · v
Power at angle θ	P = Fv cosθ

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23. One-Minute Revision 🚀

Before the JEE exam, remember this simple flow:

Force + Displacement → Work

Motion → Kinetic Energy

Net Work → Change in Kinetic Energy

W_net = ΔK

For constant force:

W = Fs cosθ

Kinetic energy:

K = ½mv²

Change in kinetic energy:

ΔK = ½m(v² − u²)

Gravity:

W_g = mg(h_i − h_f)

Spring:

U = ½kx²

Variable force:

W = ∫F dx

F-x graph:

Area = Work

Power:

P = W/t

P = F · v

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24. Final Revision Box 🔥

WORK:

W = Fs cosθ

KINETIC ENERGY:

K = ½mv²

⭐ WORK-ENERGY THEOREM:

W_net = ΔK

Therefore:

W_net = ½m(v² − u²)

GRAVITY:

W_g = mg(h_i − h_f)

SPRING:

U_s = ½kx²

VARIABLE FORCE:

W = ∫F dx

F-x GRAPH:

Area under F-x graph = Work

POWER:

P = W/t

P = F · v

🔥 GOLDEN RULE: NET WORK → CHANGE IN KINETIC ENERGY

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25. Practice Questions for JEE Main

1. A 5 kg block changes its speed from 2 m/s to 6 m/s. Find the net work done.
2. A force of 20 N acts on a body through a displacement of 5 m at an angle of 60°. Find the work done.
3. A block moves 10 m on a rough horizontal surface. An applied force does 100 J work while friction does −40 J work. Find the change in kinetic energy.
4. A body of mass 2 kg falls through a height of 5 m. Find the work done by gravity.
5. A spring of force constant k is compressed by x. Find the energy stored in it.
6. The net force on a particle varies with position. Explain how the F-x graph can be used to calculate its change in kinetic energy.

Practice Tip: In every question, first ask yourself: “Can I solve this directly using W_net = ΔK?” If yes, avoid unnecessary equations of motion.

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26. PDF Notes

You can add your detailed PDF notes below so students can revise the complete Work Energy Theorem topic offline.

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27. Frequently Asked Questions (FAQs)

Q1. What is the Work-Energy Theorem?

The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy.

W_net = ΔK

Q2. Is the Work-Energy Theorem based on net work?

Yes. The theorem uses the total or net work done by all forces acting on the object.

Q3. Can work be negative?

Yes. Work is negative when the component of force along displacement is opposite to the displacement.

Q4. Does zero work mean that force is zero?

No. A non-zero force can do zero work when it is perpendicular to the displacement.

Q5. What is the fastest formula when initial and final speeds are given?

Use:

W_net = ½m(v² − u²)

Q6. What is the work done by gravity when an object falls through height h?

The work done by gravity is:

W_g = +mgh

Q7. What is the work done by gravity when an object moves upward?

If the object moves upward through height h:

W_g = −mgh

Q8. What is the work done by a variable force?

For a variable force:

W = ∫F dx

Graphically, it is represented by the area under the F-x graph.

Q9. What is the SI unit of work?

The SI unit of work is the Joule (J).

Q10. What is the SI unit of power?

The SI unit of power is the Watt (W).

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Final Thoughts

The Work-Energy Theorem is one of the most powerful shortcuts in Class 11 Physics and JEE Main. Instead of always solving a problem using force equations, acceleration and time, you can often connect the initial and final states directly through work and kinetic energy.

The most important equation to remember is:

W_net = ΔK = ½m(v² − u²)

But remember the key word: NET.

Whenever multiple forces are acting, calculate the work done by each force, add them, and then apply the Work-Energy Theorem. For gravity, remember the height-based shortcut. For springs, remember the elastic potential energy. For variable forces, think about the F-x graph.

With these concepts clear, a large number of Work, Energy and Power questions become much easier to solve.

Master the work → connect it to kinetic energy → solve faster in JEE Main.