Work-energy theorem
The work-energy theorem states that the net work done on an object equals the change in its kinetic energy, and applies to both constant and variable forces.
Why this shows up in the exam
You need to apply this theorem to solve problems involving motion and energy changes in NEET.
How NEET tests this
Learn the idea
The net work done on a particle equals the change in its kinetic energy; the key is to compute the difference ½m(v_f²‑v_i²).
🧠 Memory hook: Work‑energy is like a bank account: the net work you deposit or withdraw equals the change in the balance (½ m v²).
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Work W = F·s (scalar product)
- Kinetic energy KE = ½ m v²
- Net work = ΔKE = KE_final – KE_initial
- For variable forces, W = ∫ F·ds
- Work done by a resisting force is negative (it removes kinetic energy)
- Mass must be in kilograms for SI units
How to approach it
- 1Write down the initial and final speeds of the object
- 2Convert the mass to kilograms
- 3Compute KE_initial = ½ m v_i² and KE_final = ½ m v_f²
- 4Find ΔKE = KE_final – KE_initial; the magnitude of work done against resistance is –ΔKE (positive value)
Worked example — watch it click
A bullet of mass 10 g leaves a rifle at an initial velocity of 1000 m/s and strikes the target at the same level with a velocity of 500 m/s. The work done in joules overcoming the resistance of air will be:
- A)375
- ✅3750
- C)5000
- D)500
The concept behind this problem
The bullet loses kinetic energy while moving through air; that loss equals the work done against air resistance, which is exactly what the question asks for.
Step by step
- 1Work against air resistance = loss in KE = ½m(v₁² - v₂²) = ½(0.01)(1000² - 500²) = 0.005(750000) = 3750 J.
Watch out
Students often forget to convert 10 g to 0.01 kg, leading to a 1000‑fold error in the work value.
Common slip-ups that cost marks
- •Using gram instead of kilogram gives a factor of 1000 error
- •Treating the loss of kinetic energy as the total kinetic energy
- •Ignoring the sign: work done by air resistance is negative, but the asked value is its magnitude
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Practise it
These are real questions from past NEET papers that test this exact idea.
A bullet of mass 10 g leaves a rifle at an initial velocity of 1000 m/s and strikes the target at the same level with a velocity of 500 m/s. The work done in joules overcoming the resistance of air will be:
Push further
More challenging3 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.
A block of mass 2 kg is pushed across a horizontal surface by a constant force of 10 N over a distance of 5 m. If the block starts from rest and its final speed is 6 m/s, what is the work done by friction?
More from Work, Energy and Power
Work and its calculation
Work is the energy transferred by a force acting over a distance, and can be calculated using the dot product, area under a force-displacement graph, or for variable and constant forces.
Conservation of energy
The law of conservation of energy states that energy cannot be created or destroyed, only transformed, including cases with energy loss and efficiency considerations.
Conservative and non-conservative forces
Conservative forces, like gravity and spring force, conserve mechanical energy, while non-conservative forces, like friction, dissipate energy as heat.
Elastic potential energy
Elastic potential energy is the energy stored in a stretched or compressed spring, proportional to the square of its displacement.
Power
Power is the rate at which work is done or energy is transferred, and can be calculated as the product of force and velocity at any instant.
Work by a Constant Force
For a constant force, work is the scalar product of force and displacement, so its sign is set by the angle between those vectors and not by force magnitude alone.