Work Power Energy Formulas For JEE 2026
Work, Energy, and Power are among the most important chapters in JEE Physics, especially for students preparing for JEE Main and Advanced. Almost every year, around 2–3 questions are asked from this topic, making it essential for scoring well in the exam. The chapter explains key concepts such as work done by constant and variable forces, the work–energy theorem, conservation of mechanical energy, power, and different types of collisions including elastic and inelastic collisions. Understanding these fundamental principles helps students build a strong conceptual base in mechanics and improves their problem-solving accuracy in the exam.
A strong grasp of Work, Energy, and Power formulas allows students to solve complex numerical problems more efficiently. Instead of relying only on lengthy kinematic calculations, many JEE problems can be solved quickly using energy conservation and work–energy relationships. For quick revision and better exam preparation, students can also refer to the JEE Mains Physics Formula PDF, which provides a compiled list of important formulas from this chapter and other key JEE topics. Regular practice using these formulas can significantly improve speed, accuracy, and confidence during the exam.
What Is Work?
In everyday language, "work" means any physical or mental effort. But in physics, work has a very specific meaning: work is done only when a force moves an object through a displacement. If you push a wall and it doesn't move, you have done zero work in the physics sense, even though you feel tired.
Work: The product of the component of force along the direction of displacement and the magnitude of the displacement. Work is a scalar quantity. SI unit: joule (J), where $$1 \text{ J} = 1 \text{ N} \cdot \text{m}$$.
Work Done by a Constant Force
When a constant force $$\vec{F}$$ acts on an object that undergoes displacement $$\vec{d}$$, the work done depends on the angle $$\theta$$ between the force and the displacement.
Work Done by a Constant Force
$$$W = Fd\cos\theta = \vec{F} \cdot \vec{d}$$$where $$F$$ = magnitude of force (N), $$d$$ = magnitude of displacement (m), $$\theta$$ = angle between $$\vec{F}$$ and $$\vec{d}$$.
Special cases:
- $$\theta = 0°$$ (force along displacement): $$W = Fd$$ (maximum positive work)
- $$\theta = 90°$$ (force perpendicular to displacement): $$W = 0$$ (no work done)
- $$\theta = 180°$$ (force opposite to displacement): $$W = -Fd$$ (negative work)
Worked Example
A person pulls a box 10 m along the floor with a force of 50 N at $$60°$$ above the horizontal. Find the work done.
$$W = Fd\cos\theta = 50 \times 10 \times \cos 60° = 50 \times 10 \times 0.5 = 250$$ J.
Worked Example
A 5 kg block slides 4 m on a surface with $$\mu_k = 0.3$$. Find the work done by friction. ($$g = 10$$ m/s$$^2$$)
Friction force: $$f_k = \mu_k mg = 0.3 \times 5 \times 10 = 15$$ N.
Friction opposes motion, so $$\theta = 180°$$:
$$W_{\text{friction}} = f_k \times d \times \cos 180° = 15 \times 4 \times (-1) = -60$$ J.
The negative sign means friction removes energy from the object.
Note: The normal force and gravity do zero work when an object moves horizontally, because they are perpendicular to the displacement ($$\theta = 90°$$). However, gravity does work when there is a vertical component of displacement.
Work Done by a Variable Force
When the force changes during the motion (for example, the force needed to stretch a spring increases as you stretch it further), we cannot simply multiply $$F \times d$$. Instead, we add up (integrate) the tiny bits of work done over tiny displacements.
Work Done by a Variable Force
$$$W = \int_{x_1}^{x_2} F(x)\, dx$$$where $$F(x)$$ is the force as a function of position $$x$$.
Graphically, the work equals the area under the $$F$$-$$x$$ curve between $$x_1$$ and $$x_2$$.
Worked Example
A force $$F = 3x^2$$ N acts on a particle. Find the work done as the particle moves from $$x = 0$$ to $$x = 4$$ m.
$$W = \int_0^4 3x^2\, dx = 3 \left[\dfrac{x^3}{3}\right]_0^4 = \left[x^3\right]_0^4 = 64 - 0 = 64$$ J.
Work Done by a Spring Force
A spring obeys Hooke's law: the restoring force is proportional to the displacement from the natural (unstretched) length. The spring force is $$F = -kx$$, where the negative sign means it always acts to restore the spring to its natural length.
Spring Constant ($$k$$): A measure of the stiffness of a spring. Higher $$k$$ means a stiffer spring. SI unit: N/m.
Work Done by a Spring
When a spring is stretched or compressed from position $$x_1$$ to $$x_2$$:
$$$W_{\text{spring}} = -\frac{1}{2}k x_2^2 + \frac{1}{2}k x_1^2 = \frac{1}{2}k(x_1^2 - x_2^2)$$$From natural length ($$x_1 = 0$$) to extension $$x$$:
$$$W_{\text{spring}} = -\frac{1}{2}kx^2$$$Work done by an external agent to stretch the spring from natural length to $$x$$:
$$$W_{\text{external}} = +\frac{1}{2}kx^2$$$Worked Example
A spring with $$k = 200$$ N/m is compressed by 0.1 m. Find the work done by the spring force when it returns to its natural length.
$$W = \frac{1}{2}k(x_1^2 - x_2^2) = \frac{1}{2} \times 200 \times (0.1^2 - 0^2) = \frac{1}{2} \times 200 \times 0.01 = 1$$ J.
The spring does positive work as it pushes the object back (spring force and displacement are in the same direction).
Kinetic Energy Formula
A moving object has the ability to do work simply because it is moving — a moving hammer drives a nail, a moving ball knocks down pins. This ability to do work due to motion is called kinetic energy.
Kinetic Energy (KE): The energy possessed by an object due to its motion.
Kinetic Energy
$$$KE = \frac{1}{2}mv^2$$$where $$m$$ = mass (kg), $$v$$ = speed (m/s).
Key properties:
- Kinetic energy is always $$\geq 0$$ (it can never be negative).
- It is a scalar quantity.
- It depends on the square of speed — doubling the speed quadruples the KE.
- SI unit: joule (J).
Worked Example
Find the kinetic energy of a 2 kg ball moving at 5 m/s.
$$KE = \frac{1}{2} \times 2 \times 5^2 = \frac{1}{2} \times 2 \times 25 = 25$$ J.
Work–Energy Theorem
There is a beautiful connection between work and kinetic energy: the net work done on an object equals the change in its kinetic energy. This is one of the most powerful results in mechanics.
Work–Energy Theorem
$$$W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$$where $$u$$ = initial speed, $$v$$ = final speed.
In words: the total work done by all forces on an object equals the change in its kinetic energy.
Worked Example
A 4 kg block initially at rest is pushed by a net force of 20 N over 10 m. Find the final speed.
$$W_{\text{net}} = Fd = 20 \times 10 = 200$$ J.
Using the work–energy theorem:
$$200 = \frac{1}{2} \times 4 \times v^2 - 0$$
$$v^2 = \dfrac{200 \times 2}{4} = 100$$
$$v = 10$$ m/s.
Worked Example
A 2 kg ball moving at 10 m/s is brought to rest by a force over 5 m. Find the retarding force.
$$W_{\text{net}} = \Delta KE = \frac{1}{2}(2)(0)^2 - \frac{1}{2}(2)(10)^2 = 0 - 100 = -100$$ J.
$$W = Fd\cos 180° = -F \times 5$$
$$-100 = -5F \Rightarrow F = 20$$ N.
Potential Energy Formulas — Gravitational and Elastic
Some forms of energy are "stored" and can be released later. A ball held at a height has energy due to its position (it will gain speed if dropped). A compressed spring has energy due to its deformation (it will push an object when released). This stored energy is called potential energy.
Potential Energy: Energy stored in an object due to its position or configuration. It is associated with conservative forces (gravity, spring force, electrostatic force).
Gravitational Potential Energy
When you lift an object to a height $$h$$ above a reference level, you do work against gravity. This work gets stored as gravitational potential energy.
Gravitational Potential Energy
$$$PE = mgh$$$where $$m$$ = mass (kg), $$g$$ = acceleration due to gravity (m/s$$^2$$), $$h$$ = height above the reference level (m).
Key points:
- $$h$$ is measured from a chosen reference level (often the ground). The choice of reference doesn't affect energy differences.
- PE can be positive, negative, or zero depending on the reference level.
- Work done by gravity: $$W_{\text{gravity}} = -\Delta PE = mgh_1 - mgh_2 = -mg(h_2 - h_1)$$.
Worked Example
A 3 kg book is lifted from the floor to a shelf 2 m high. Find the change in gravitational PE. ($$g = 10$$ m/s$$^2$$)
$$\Delta PE = mgh = 3 \times 10 \times 2 = 60$$ J.
This 60 J of work was done by the person lifting the book and is now stored as gravitational PE.
Elastic Potential Energy
When a spring is stretched or compressed from its natural length, energy is stored in it. This is elastic potential energy.
Elastic Potential Energy
$$$PE_{\text{spring}} = \frac{1}{2}kx^2$$$where $$k$$ = spring constant (N/m), $$x$$ = displacement from natural length (m).
This energy is always $$\geq 0$$ (whether stretched or compressed).
Worked Example
A spring with $$k = 500$$ N/m is compressed by 0.2 m. Find the stored elastic PE.
$$PE = \frac{1}{2} \times 500 \times (0.2)^2 = \frac{1}{2} \times 500 \times 0.04 = 10$$ J.
Conservation of Energy Formulas
Energy can change from one form to another (kinetic to potential, potential to kinetic, etc.), but the total energy of an isolated system remains constant. This is one of the most fundamental principles in all of physics.
Conservative Force: A force for which the work done depends only on the initial and final positions, not on the path taken. Examples: gravity, spring force, electrostatic force.
Non-Conservative Force: A force for which the work done depends on the path. Example: friction, air resistance.
Conservation of Mechanical Energy
If only conservative forces act (no friction, no air resistance):
$$$KE_1 + PE_1 = KE_2 + PE_2$$$ $$$\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2$$$If non-conservative forces also act:
$$$KE_1 + PE_1 + W_{\text{nc}} = KE_2 + PE_2$$$where $$W_{\text{nc}}$$ = work done by non-conservative forces (usually negative for friction).
Worked Example
A ball is dropped from 20 m. Find its speed just before hitting the ground. ($$g = 10$$ m/s$$^2$$)
Using conservation of energy (taking ground as reference):
$$mgh = \frac{1}{2}mv^2$$
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20$$ m/s.
Notice: the mass cancels out — all objects fall with the same speed (ignoring air resistance).
Worked Example
A 2 kg block slides down a 5 m high frictionless ramp. Find the speed at the bottom.
$$mgh = \frac{1}{2}mv^2$$
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10$$ m/s.
This result is independent of the shape of the ramp (steep, gentle, curved) — only the height matters for a frictionless surface.
Worked Example
A 5 kg block slides down a 3 m high rough incline ($$\mu_k = 0.2$$, incline angle $$30°$$). Find the speed at the bottom. ($$g = 10$$ m/s$$^2$$)
Length of incline: $$L = \dfrac{h}{\sin\theta} = \dfrac{3}{\sin 30°} = \dfrac{3}{0.5} = 6$$ m.
Normal force: $$N = mg\cos 30° = 5 \times 10 \times 0.866 = 43.3$$ N.
Friction force: $$f_k = \mu_k N = 0.2 \times 43.3 = 8.66$$ N.
Work done by friction: $$W_f = -f_k L = -8.66 \times 6 = -51.96$$ J.
Using the energy equation with friction:
$$mgh + W_f = \frac{1}{2}mv^2$$
$$5 \times 10 \times 3 + (-51.96) = \frac{1}{2} \times 5 \times v^2$$
$$150 - 51.96 = 2.5 v^2$$
$$v^2 = \dfrac{98.04}{2.5} = 39.22$$
$$v = 6.26$$ m/s.
Tip: Energy conservation is often faster than using $$F = ma$$ and kinematics, especially when the path is curved or when you only need the final speed (not the time).
Power Formulas
Two motors might do the same amount of work (say, lifting a 100 kg load by 10 m), but one might do it in 5 seconds and the other in 50 seconds. The faster motor is more "powerful." Power measures how quickly work is done.
Power: The rate of doing work, or the rate of energy transfer. SI unit: watt (W), where $$1 \text{ W} = 1 \text{ J/s}$$.
Power Formulas
Average power:
$$$P_{\text{avg}} = \frac{W}{t} = \frac{\text{Total work done}}{\text{Total time taken}}$$$Instantaneous power:
$$$P = \frac{dW}{dt} = \vec{F} \cdot \vec{v} = Fv\cos\theta$$$where $$\vec{v}$$ is the instantaneous velocity.
If force is along the direction of velocity ($$\theta = 0$$):
$$$P = Fv$$$Other units: 1 horsepower (hp) $$= 746$$ W. 1 kW $$= 1000$$ W.
Energy unit: 1 kilowatt-hour (kWh) $$= 3.6 \times 10^6$$ J.
Worked Example
A crane lifts a 200 kg load by 15 m in 10 seconds. Find the power. ($$g = 10$$ m/s$$^2$$)
Work done $$= mgh = 200 \times 10 \times 15 = 30000$$ J.
Power $$= \dfrac{W}{t} = \dfrac{30000}{10} = 3000$$ W $$= 3$$ kW.
Worked Example
A car engine delivers constant power $$P = 40$$ kW. If the car has mass 1000 kg, find the maximum speed on a level road where air resistance is $$F_{\text{air}} = 800$$ N.
At maximum speed, acceleration $$= 0$$, so the engine force equals the air resistance:
$$P = F_{\text{air}} \times v_{\max}$$
$$40000 = 800 \times v_{\max}$$
$$v_{\max} = 50$$ m/s $$= 180$$ km/h.
Tip: When a vehicle moves at constant speed, all the engine power goes into overcoming friction/air resistance. At maximum speed, $$P = F_{\text{resistance}} \times v_{\max}$$. This is a very common JEE question.
Collision Formulas — Elastic and Inelastic
When two objects strike each other, we call it a collision. During a collision, the objects exert large forces on each other for a short time. Understanding collisions is important because momentum is always conserved, but kinetic energy may or may not be.
Collision: A short-duration interaction between two bodies where they exert strong forces on each other. The external forces (like gravity) are negligible compared to the collision forces during the brief interaction.
Conservation of Momentum in Collisions
Since the collision forces are internal to the two-body system (by Newton's third law, they are equal and opposite), the net external force is approximately zero during the collision. Therefore, total momentum is conserved in all types of collisions.
Conservation of Momentum
$$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$$where $$u_1, u_2$$ = velocities before collision, $$v_1, v_2$$ = velocities after collision.
This applies to all collisions (elastic, inelastic, perfectly inelastic).
Types of Collisions
Elastic vs Inelastic Collisions
| Property | Elastic | Perfectly Inelastic |
|---|---|---|
| Momentum | Conserved | Conserved |
| Kinetic energy | Conserved | Not conserved (max loss) |
| Bodies after | Separate | Stick together |
| Coeff. of restitution | $$e = 1$$ | $$e = 0$$ |
Partially inelastic (most real collisions): momentum conserved, some KE lost, $$0 < e < 1$$.
Coefficient of Restitution
Coefficient of Restitution ($$e$$): A measure of the "bounciness" of a collision. It is the ratio of the relative speed of separation to the relative speed of approach.
Coefficient of Restitution
$$$e = \frac{\text{Relative speed of separation}}{\text{Relative speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$$$where $$u_1, u_2$$ are initial velocities and $$v_1, v_2$$ are final velocities (in 1D, with appropriate signs).
- $$e = 1$$: perfectly elastic collision.
- $$e = 0$$: perfectly inelastic collision (bodies stick together).
- $$0 < e < 1$$: partially inelastic (most real collisions).
Elastic Collision Formulas (1D)
For a head-on elastic collision between two masses $$m_1$$ and $$m_2$$, both momentum and kinetic energy are conserved. Solving these simultaneously gives:
Elastic Collision — Final Velocities
$$$v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}$$$ $$$v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}$$$Special cases (target at rest, $$u_2 = 0$$):
- $$m_1 = m_2$$: $$v_1 = 0$$, $$v_2 = u_1$$ (complete transfer of velocity).
- $$m_1 \gg m_2$$: $$v_1 \approx u_1$$, $$v_2 \approx 2u_1$$ (light ball bounces off fast).
- $$m_1 \ll m_2$$: $$v_1 \approx -u_1$$, $$v_2 \approx 0$$ (bounces back, heavy ball barely moves).
Worked Example
A 2 kg ball moving at 6 m/s collides elastically head-on with a 4 kg ball at rest. Find the velocities after collision.
$$v_1 = \dfrac{(2 - 4)(6) + 2(4)(0)}{2 + 4} = \dfrac{-12}{6} = -2$$ m/s (bounces back).
$$v_2 = \dfrac{(4 - 2)(0) + 2(2)(6)}{2 + 4} = \dfrac{24}{6} = 4$$ m/s (moves forward).
Verification:
Momentum before: $$2 \times 6 + 4 \times 0 = 12$$ kg·m/s.
Momentum after: $$2 \times (-2) + 4 \times 4 = -4 + 16 = 12$$ kg·m/s. ✓
KE before: $$\frac{1}{2}(2)(36) = 36$$ J.
KE after: $$\frac{1}{2}(2)(4) + \frac{1}{2}(4)(16) = 4 + 32 = 36$$ J. ✓
Perfectly Inelastic Collision
In a perfectly inelastic collision, the two objects stick together after impact and move as one combined mass. This results in the maximum possible loss of kinetic energy (while still conserving momentum).
Perfectly Inelastic Collision
After the collision, both bodies move together with velocity:
$$$v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$$$Loss of kinetic energy:
$$$\Delta KE = \frac{1}{2} \cdot \frac{m_1 m_2}{m_1 + m_2} \cdot (u_1 - u_2)^2$$$Worked Example
A 3 kg ball moving at 8 m/s collides with a 5 kg ball at rest and they stick together. Find the common velocity and the kinetic energy lost.
$$v = \dfrac{3 \times 8 + 5 \times 0}{3 + 5} = \dfrac{24}{8} = 3$$ m/s.
KE before $$= \frac{1}{2}(3)(64) = 96$$ J.
KE after $$= \frac{1}{2}(8)(9) = 36$$ J.
KE lost $$= 96 - 36 = 60$$ J.
Using the formula: $$\Delta KE = \frac{1}{2} \times \dfrac{3 \times 5}{3 + 5} \times (8 - 0)^2 = \frac{1}{2} \times \dfrac{15}{8} \times 64 = 60$$ J. ✓
Note: In a perfectly inelastic collision, kinetic energy is not destroyed — it is converted into heat, sound, and deformation energy. Total energy is always conserved; only kinetic energy is lost.
Collision with a Wall or Floor
When a ball bounces off a rigid wall or floor, the wall has effectively infinite mass and doesn't move. The coefficient of restitution determines the speed after bouncing.
Ball Bouncing Off a Surface
If a ball hits a surface with speed $$u$$ and bounces back with speed $$v$$:
$$$v = eu$$$where $$e$$ = coefficient of restitution.
Height after bounce: If dropped from height $$h$$, it bounces to:
$$$h' = e^2 h$$$After $$n$$ bounces: $$h_n = e^{2n} h$$
Worked Example
A ball is dropped from 10 m and the coefficient of restitution with the floor is 0.8. Find the height after the first bounce and the speed just after the first bounce. ($$g = 10$$ m/s$$^2$$)
Speed just before hitting floor: $$u = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = \sqrt{200}$$ m/s.
Speed just after bounce: $$v = eu = 0.8\sqrt{200}$$ m/s.
Height after bounce: $$h' = e^2 h = (0.8)^2 \times 10 = 0.64 \times 10 = 6.4$$ m.
Summary of Key Formulas
Work, Energy & Power Formula Sheet
| Quantity | Formula |
|---|---|
| Work (constant force) | $$W = Fd\cos\theta$$ |
| Work (variable force) | $$W = \int F\,dx$$ |
| Spring work (to stretch by $$x$$) | $$W = \frac{1}{2}kx^2$$ |
| Kinetic energy | $$KE = \frac{1}{2}mv^2$$ |
| Work–energy theorem | $$W_{\text{net}} = \Delta KE$$ |
| Gravitational PE | $$PE = mgh$$ |
| Elastic PE | $$PE = \frac{1}{2}kx^2$$ |
| Energy conservation | $$KE_1 + PE_1 = KE_2 + PE_2$$ |
| Average power | $$P = W/t$$ |
| Instantaneous power | $$P = Fv\cos\theta$$ |
| Perfectly inelastic velocity | $$v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$$ |
| Coefficient of restitution | $$e = \frac{v_2 - v_1}{u_1 - u_2}$$ |
| Bounce height | $$h' = e^2 h$$ |
Tip: Energy methods are the fastest way to solve most JEE mechanics problems. Use energy conservation when you need final speeds or heights. Use the work–energy theorem when friction or non-conservative forces are present. Use $$P = Fv$$ for problems involving engines and vehicles at constant speed.
Work Power Energy Formulas For JEE 2026: Conclusion
The chapter on Work, Power, and Energy is one of the core topics in JEE Physics and forms the foundation for many mechanics problems. Concepts such as work done by forces, kinetic and potential energy, conservation of energy, and power help students understand how energy is transferred and transformed in physical systems. Mastering the Work Power Energy Formulas For JEE 2026 is essential for solving numerical questions efficiently in both JEE Main and JEE Advanced.
For effective revision and exam preparation, students should regularly practice problems and review important formulas. Using a structured JEE Mains Physics Formula PDF can make revision faster and more organized, helping students remember key equations during the exam. With consistent practice and a clear understanding of these formulas, students can improve their speed, accuracy, and confidence in solving JEE Physics questions.
