WORK, ENERGY AND POWER | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
WORK, ENERGY AND POWER – this guide gives you a concise, exam-ready overview of WORK, ENERGY AND POWER from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
5.1.1 The Scalar Product
This section introduces the scalar product (also called the dot product) of two vectors, a fundamental mathematical operation used in defining work. Given two vectors A and B, their scalar product A · B is defined as the product of their magnitudes and the cosine of the angle θ between them: A · B = A B cos θ. This product is a scalar quantity, meaning it has magnitude but no direction. Geometrically, the scalar product can be interpreted as the magnitude of one vector multiplied by the projection of the other vector along it. The scalar product is commutative (A · B = B · A) and distributive over vector addition (A · (B + C) = A · B + A · C). For unit vectors along coordinate axes (i, j, k), the scalar product is 1 if the vectors are the same and 0 if they are perpendicular. The scalar product of two vectors expressed in components is the sum of the products of their corresponding components: A · B = A_x B_x + A_y B_y + A_z B_z. This leads to the formula for the magnitude of a vector A as A = √(A_x² + A_y² + A_z²). The section includes a worked example calculating the angle between two vectors and the projection of one vector on another, illustrating the practical use of the scalar product.
📊 Diagram: Fig. 5.1(a) shows two vectors A and B with angle θ between them; the scalar product is A B cos θ. Fig. 5.1(b) shows projection of B onto A as B cos θ. Fig. 5.1(c) shows projection of A onto B as A cos θ.
🧪 Activity: No specific activity, but the section encourages proving properties of scalar product as exercises.
🔗 Connection: Provides mathematical foundation for defining work in the next section 5.3.
Frequently asked questions
5.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Solution: (a) Work done by the man in lifting the bucket is positive because the force applied by the man and the displacement of the bucket are in the same direction (upwards).
(b) Work done by gravitational force is negative because gravity acts downward while the displacement is upward.
(c) Work done by friction on a body sliding down an inclined plane is negative because friction opposes the motion.
(d) Work done by an applied force on a body moving with uniform velocity on a rough horizo
5.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.
Given: Mass, m = 2 kg Applied force, F = 7 N Coefficient of kinetic friction, μ = 0.1 Time, t = 10 s
Step 1: Calculate friction force: Normal force, N = mg = 2 × 9.8 = 19.6 N Friction force, f = μN = 0.1 × 19.6 = 1.96 N
Step 2: Calculate net force: Net force, F_net = Applied force - friction = 7 - 1.96 = 5.04 N
Step 3: Calculate acceleration: a = F_net / m = 5.04 / 2 = 2.52 m/s²
Step 4: Calculate displacement in 10 s: s = ut + 0.5 a t² = 0 + 0.5 × 2.52 × 100 = 126 m
(a) Work done by applied
5.3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Solution: For each potential energy graph:
- The particle cannot be found in regions where potential energy V(x) > total energy E, because kinetic energy K = E - V(x) would be negative, which is impossible.
- Minimum total energy is the lowest point on the potential energy curve.
Physical contexts:
- Parabolic potential (harmonic oscillator) corresponds to mass on a spring.
- Potential wells correspond to bound states in quantum mechanics or classical particles trapped in a potential.
- Potenti
5.4 The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = kx^2 / 2$ , where $k$ is the force constant of the oscillator. For $k = 0.5\mathrm{Nm}^{-1}$ , the graph of $V(x)$ versus $x$ is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches $x = \pm 2\mathrm{m}$ .
Given: Potential energy, V(x) = (k x^2)/2 k = 0.5 N/m Total energy, E = 1 J
At turning points, kinetic energy K = 0, so total energy E = potential energy V(x).
Calculate V(x) at x = 2 m: V(2) = 0.5 × (2)^2 / 2 = 0.5 × 4 / 2 = 1 J
Since V(2) = E, the particle cannot go beyond x = ±2 m because beyond this point potential energy would exceed total energy, making kinetic energy negative.
Hence, the particle turns back at x = ±2 m.
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