Class 11 — Work, Energy and Power

Chapter Overview & Weightage

Work, Energy and Power is one of the most scoring chapters in CBSE Class 11 Physics. The questions are predictable, the formulas are few, and the marking scheme is generous to candidates who write neat steps. We typically see one 5-mark question and one 2-3 mark question every year.

Year	Marks Allotted	Question Type
2024	7	Numerical + theory
2023	5	Derivation + numerical
2022	6	Conservation problem
2021	5	Variable force, integration
2020	7	Power + collisions combo

Average weightage: 6 marks per board paper. With Class 12 not testing this chapter, all your investment pays off in one shot during Class 11 boards.

Key Concepts You Must Know

Work done by constant force: $W = \vec F \cdot \vec d = Fd\cos\theta$ .
Work done by variable force: $W = \int F\,dx$ .
Work-energy theorem: $W_{\text{net}} = \Delta KE$ .
Kinetic energy: $KE = \tfrac{1}{2}mv^2$ .
Potential energy: gravitational $U = mgh$ , spring $U = \tfrac{1}{2}kx^2$ .
Conservation of mechanical energy: $KE + PE =$ constant for conservative forces.
Power: $P = W/t$ , instantaneous $P = \vec F \cdot \vec v$ .
Collisions: elastic conserves both momentum and KE; inelastic conserves only momentum.

Important Formulas

$W = Fd\cos\theta \quad\text{(constant force)}$

$W = \int_{x_1}^{x_2} F(x)\,dx \quad\text{(variable force)}$

$KE = \tfrac{1}{2}mv^2$

$W_{\text{net}} = \Delta KE$

Use these when force is constant on a straight line. The integral form is needed when force depends on position.

$P_{\text{avg}} = \frac{W}{t}, \quad P_{\text{inst}} = \vec F \cdot \vec v$

The instantaneous form is essential when force or velocity changes — common in pulley and inclined-plane setups.

For elastic collision, masses $m_1, m_2$ with initial velocities $u_1, u_2$ :

$v_1 = \frac{(m_1 - m_2)u_1 + 2 m_2 u_2}{m_1 + m_2}$

$v_2 = \frac{(m_2 - m_1)u_2 + 2 m_1 u_1}{m_1 + m_2}$

For perfectly inelastic, both masses stick: $v = (m_1 u_1 + m_2 u_2)/(m_1 + m_2)$ .

Solved Previous Year Questions

PYQ 1 (CBSE 2024)

A force $\vec F = (3\hat i + 4\hat j)$ N displaces an object by $\vec d = (5\hat i + 2\hat j)$ m. Calculate the work done.

$W = \vec F \cdot \vec d = (3)(5) + (4)(2) = 15 + 8 = 23\text{ J}$ .

PYQ 2 (CBSE 2023)

A spring of force constant $k = 200\text{ N/m}$ is compressed by $0.1\text{ m}$ . Find the energy stored. If a $0.5\text{ kg}$ ball is released, find its speed when the spring returns to natural length.

$U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.01) = 1\text{ J}$ .

By energy conservation, this becomes KE: $\tfrac{1}{2}(0.5)v^2 = 1 \implies v^2 = 4 \implies v = 2\text{ m/s}$ .

PYQ 3 (CBSE 2022)

A motor lifts a $1000\text{ kg}$ load to a height of $20\text{ m}$ in $40\text{ s}$ . Find the power.

$W = mgh = 1000 \times 10 \times 20 = 2 \times 10^5\text{ J}$ . $P = W/t = 2\times 10^5 / 40 = 5000\text{ W} = 5\text{ kW}$ .

Difficulty Distribution

Sub-topic	Easy	Medium	Hard
Work definition	60%	30%	10%
Energy theorems	30%	50%	20%
Power problems	50%	40%	10%
Collisions	20%	50%	30%

Easy questions test direct formula recall; medium combines two concepts; hard ties in calculus or 2D collisions.

Expert Strategy

Toppers always start collision problems by writing momentum and KE conservation as separate equations, then solve together. Substituting half-derived expressions back is where errors creep in.

For variable force problems, always sketch $F$ vs $x$ first. If the graph is a straight line, the area is a triangle — no integration needed.

Memorise that for elastic collisions, equal masses exchange velocities. This shortcut clears every “billiard ball” type MCQ in seconds.

Common Traps

Confusing average power and instantaneous power. The 5-mark question often gives a force varying with time and asks for instantaneous $P$ at $t = 5\text{ s}$ — not the average.

Forgetting that work done by friction is always negative when the body slides forward. Students sometimes write $W_f = +\mu N d$ , missing the cosine of $180°$ .

Treating perfectly inelastic collisions as conserving KE. They do not — only momentum is conserved. The KE loss equals the energy that goes into heat or deformation.