Work, Energy and Power is a chapter where JEE rarely asks you to simply recall a formula – it asks you to recognize which formula applies to a situation that’s often dressed up to look more complicated than it is. The fastest way to build that recognition is to work through real, fully-solved problems rather than just re-reading formula definitions. Here are six problems spanning the full breadth of this chapter, each solved completely, with the reasoning behind every step made explicit.
Problem 1: Work Done by a Variable Force
Question: A force F = (3x² + 2x) N acts on a particle along the x-axis. Find the work done in moving the particle from x = 1 m to x = 3 m.
Solution: Since the force varies with position, the work done isn’t a simple F×d calculation – it requires integration:
W = ∫F dx = ∫(3x² + 2x) dx from 1 to 3
W = [x³ + x²] from 1 to 3 = (27 + 9) – (1 + 1) = 36 – 2 = 34 J
The key recognition here is spotting that “F depends on x” is the signal to integrate rather than multiply directly – a distinction covered in more depth on the work done by variable force page, which is worth revisiting if this integration step felt unfamiliar rather than automatic.
Problem 2: Work-Energy Theorem With a Variable Force
Question: A 2 kg block starts from rest and is acted on by the same force F = (3x² + 2x) N from Problem 1. Find its speed at x = 3 m.
Solution: Rather than solving Newton’s second law directly (which would require dealing with acceleration as a function of position – genuinely messy), apply the work-energy theorem instead: the net work done equals the change in kinetic energy.
We already found W = 34 J from Problem 1. Since the block starts from rest, initial KE = 0.
34 = ½(2)v² – 0
v² = 34
v = √34 ≈ 5.83 m/s
This is exactly the kind of shortcut the work-energy theorem for variable force page is built around – recognizing that you almost never need to solve the messy differential equation of motion directly when the work-energy theorem gives you a two-line path to the same answer.
Problem 3: Power Delivered by a Variable Force
Question: Using the same scenario as Problems 1 and 2, find the instantaneous power delivered by the force at x = 3 m.
Solution: Power is the rate of work done, P = F·v. We already know both quantities at x = 3 m:
F = 3(3)² + 2(3) = 27 + 6 = 33 N
v = √34 m/s (from Problem 2)
P = F × v = 33 × √34 ≈ 192.4 W
Notice how Problems 1 through 3 build on each other – this is a very common JEE pattern: a single physical scenario tested across work, then velocity via energy, then power, all using results carried forward from the previous part. The Work, Energy and Power overview page ties these three quantities together conceptually, which is useful context before attempting multi-part numericals like this one.
Problem 4: Elastic Potential Energy in a Spring
Question: A spring with force constant k = 200 N/m is compressed by 0.1 m. It’s then used to launch a 0.5 kg block. Find the maximum speed of the block, assuming no energy loss to friction.
Solution: The energy stored in the compressed spring converts entirely into kinetic energy of the block, since there’s no friction to account for.
Elastic PE = ½kx² = ½(200)(0.1)² = ½(200)(0.01) = 1 J
Setting this equal to the block’s kinetic energy:
1 = ½(0.5)v²
v² = 4
v = 2 m/s
The core move here is recognizing that “spring launches a block, no friction” is a conservation-of-energy setup, not a force-and-acceleration problem – trying to solve this through F = ma and kinematics would be far more complicated than simply equating the two energy forms directly. The potential energy of a spring page covers this formula’s derivation in more detail.
Problem 5: Conservation of Mechanical Energy on an Incline
Question: A 1 kg block slides down a frictionless incline from a height of 5 m. Find its speed at the bottom.
Solution: Since the incline is frictionless, mechanical energy is conserved – all gravitational potential energy at the top converts to kinetic energy at the bottom, regardless of the incline’s angle.
mgh = ½mv²
Notice the mass cancels out entirely:
gh = ½v²
v² = 2gh = 2(10)(5) = 100
v = 10 m/s
This is a frequently tested JEE trap in disguise – students often assume the incline’s angle matters for the final speed, and waste time trying to incorporate it, when for a frictionless incline, only the height matters, not the angle or the path length. This principle is explored in full on the conservation of mechanical energy page, which is worth reviewing specifically for this exact misconception.
Problem 6: Kinetic Energy Loss in a Collision
Question: A 2 kg ball moving at 6 m/s collides head-on with a stationary 1 kg ball. After the collision, they stick together. Find the kinetic energy lost in the collision.
Solution: This is a perfectly inelastic collision (the two objects stick together), so first find the common velocity using conservation of momentum:
m₁u₁ + m₂u₂ = (m₁+m₂)v
(2)(6) + (1)(0) = (3)v
12 = 3v
v = 4 m/s
Initial KE = ½(2)(6)² + ½(1)(0)² = 36 J
Final KE = ½(3)(4)² = 24 J
KE lost = 36 – 24 = 12 J
The key idea worth internalizing: momentum is always conserved in a collision, but kinetic energy is only conserved in elastic collisions – perfectly inelastic collisions like this one always lose some kinetic energy, converted into heat, sound, or deformation. The full classification of collision types and how JEE tests the distinction between them is covered on the collisions page.
The Pattern Across All Six Problems
Look back across these six numericals and a clear pattern emerges: JEE rarely asks you to plug numbers into a single formula in isolation. Problem 1 needed integration before you could even reach a number. Problems 2 and 3 reused that number twice more. Problems 4, 5, and 6 all hinged on correctly identifying which conservation principle applied – energy conservation for the spring and incline, but momentum conservation (with energy loss) for the collision. Recognizing which principle governs a given scenario, before you touch a single formula, is the actual skill this chapter is testing.
For a broader set of numericals covering the rest of your Mechanics syllabus alongside Work, Energy and Power, the Mechanics problem-solving page is a natural next step, and practicing against real exam conditions through JEE Main previous year question papers will show you how often each of these six problem types actually appears on the paper itself.
Frequently Asked Questions
Is it necessary to memorize the work-energy theorem separately from Newton’s second law, or are they the same thing?
They’re derived from the same foundation but solve problems differently – the work-energy theorem is often faster for variable forces or when you only need speed, not full details of the motion, whereas Newton’s second law is more direct for constant-force, constant-acceleration scenarios.
Why does mass cancel out in the frictionless incline problem but not in the collision problem?
In the incline problem, both potential and kinetic energy scale identically with mass, so it cancels naturally. In the collision problem, the two objects have different masses interacting with each other, so mass can’t cancel the same way – it directly determines the shared final velocity.
How do I know whether a collision is elastic or perfectly inelastic in a JEE question?
The question usually states it directly (“stick together” signals perfectly inelastic; “kinetic energy is conserved” or “elastic collision” signals elastic) – if neither is stated explicitly, check whether the objects move independently after collision (elastic or partially inelastic) or together (perfectly inelastic).
Work, Energy and Power numericals reward pattern recognition over memorized formulas. Once you can look at a problem and immediately identify which conservation principle governs it – energy, momentum, or both – the actual arithmetic becomes the easy part, exactly as it did across all six problems above.
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