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.
Why this shows up in the exam
NEET tests your ability to compute work in different scenarios and interpret force-displacement relationships.
How NEET tests this
Learn the idea
Work is the scalar product of force and displacement – the amount of energy transferred when a force acts through a distance. The key insight is that only the component of force along the displacement matters, so for a variable force you integrate that component over the path.
🧠 Memory hook: Think of pushing a sled: only the push in the direction you slide the sled (F cosθ) actually moves it – work = that push × distance.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Work = F·s = |F| |s| cosθ (NCERT)
- For a constant force: W = F s cosθ
- For a variable force: W = ∫ F·dr = ∫ F cosθ ds
- If force ⟂ displacement, work is zero
- Unit of work is joule (1 J = 1 N·m)
How to approach it
- 1Identify whether the force is constant or expressed as a function of the coordinate along the path
- 2Resolve the force into the component parallel to the displacement (multiply by cosθ)
- 3If constant, use W = F s cosθ; if variable, write F as a function of the coordinate and evaluate W = ∫ F cosθ ds over the given limits
- 4Insert the limits of motion and compute the integral or simple product
- 5State the answer with correct sign and unit
Worked example — watch it click
A force F = 20 + 10y acts on a particle in y-direction where F is in newtons and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is:
- A)5 J
- ✅25 J
- C)20 J
- D)30 J
The concept behind this problem
The example requires you to treat a linearly increasing force as a function of y and integrate it from y=0 to y=1, directly applying the variable‑force work formula.
Step by step
- 1Work = ∫ F dy from y=0 to y=1.
- 2F = 20 + 10y. ∫(20 + 10y) dy = 20y + 5y² from 0 to 1 = 20×1 + 5×1 = 25 J.
Watch out
Students often take the force as constant at its initial value (20 N) and compute 20 × 1 = 20 J, missing the contribution of the increasing part.
Common slip-ups that cost marks
- •Treating the whole force instead of its component along the motion
- •Using the total displacement without integrating a varying force
- •Ignoring the sign – a force opposite to motion gives negative work
🌟 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 force F = 20 + 10y acts on a particle in y-direction where F is in newtons and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is:
More from Work, Energy and Power
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.
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.
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.