Class 11 Physics: Work, Energy and Power | Physics Q&A

CBSE Class 11 › Unit IV: Work, Energy & Power

Work, Energy and Power

NCERT Chapter 5 • Work-Energy Theorem, Potential Energy & Collisions

NCERT 2025–26 Unit IV • ~8 Marks JEE Main • 1-2 Questions

Roller coaster loop demonstrating conservation of mechanical energy. — Energy transforms but is never lost: The essence of Mechanics.

1. Work and Scalar Product

In physics, Work is said to be done only when a force produces a displacement in the direction of the force. We use the Scalar (Dot) Product to define it.

1.1 Scalar Product

\vec{A} \cdot \vec{B} = AB \cos \theta

$\theta < 90^\circ$ : Work is Positive.
$\theta = 90^\circ$ : Work is Zero (e.g., Centripetal force).
$\theta > 90^\circ$ : Work is Negative (e.g., Friction).

1.2 Work Done by a Force

Constant Force: $W = \vec{F} \cdot \vec{d} = F d \cos \theta$ .

Force-Displacement graph showing work done as area under the curve. — For variable forces, Work is the area under the F-x curve.

Variable Force: If force changes with position, we integrate over the limits:

W = \int_{x_i}^{x_f} F(x) \, dx

Problem 1: A force $\vec{F} = (3\hat{i} + 4\hat{j})$ N acts on a body, displacing it from origin to point $(2\text{m}, 3\text{m})$ . Find Work done.

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Displacement $\vec{d} = (2-0)\hat{i} + (3-0)\hat{j} = 2\hat{i} + 3\hat{j}$ m.
$W = \vec{F} \cdot \vec{d} = (3)(2) + (4)(3) = 6 + 12 = 18$ J.

2. Kinetic Energy & Work-Energy Theorem

Kinetic Energy ( $K$ ) is the energy possessed by a body by virtue of its motion. $K = \dfrac{1}{2}mv^2$ .

The Work-Energy Theorem

The work done by the net force on a particle equals the change in its kinetic energy.

W_{net} = \Delta K = K_f - K_i

Derivation: Work-Energy Theorem (Variable Force) 3 Marks

Step 1: Kinetic Energy Rate of Change
Rate of change of K is $\dfrac{dK}{dt} = \dfrac{d}{dt}(\dfrac{1}{2}mv^2) = m v \dfrac{dv}{dt}$ .

Step 2: Apply Newton’s Law
Since $F = m \dfrac{dv}{dt}$ , we get $\dfrac{dK}{dt} = F v = F \dfrac{dx}{dt}$ .

Step 3: Integrate
$dK = F dx$ . Integrating from $x_i$ to $x_f$ :
$\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F dx$ .

Step 4: Final Result
$K_f - K_i = W$ .

Problem 2: A bullet of mass 10 g leaves a rifle with velocity 400 m/s. Calculate its Kinetic Energy.

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$m = 0.01$ kg, $v = 400$ m/s.
$K = \dfrac{1}{2}mv^2 = 0.5 \times 0.01 \times (400)^2$
$K = 0.005 \times 160000 = 800$ J.

3. Potential Energy

Potential Energy ( $U$ ) is the stored energy by virtue of position or configuration. It is defined only for Conservative Forces (like Gravity, Spring Force).

F = -\dfrac{dU}{dx}

Gravitational P.E: $U = mgh$

Note:

The value of potential energy depends on the choice of reference point (where $U = 0$ ). However, the change in potential energy ( $\Delta U$ ) is independent of this choice and physically meaningful.

Why friction has no potential energy: The work done by kinetic friction depends on the total path length, not just start and end points. Such forces are called non-conservative.

Spring system and its corresponding parabolic potential energy curve. — The energy stored in a spring depends on the square of its displacement.

Potential Energy of a Spring

For a spring obeying Hooke’s Law ( $F_s = -kx$ ), the work done against the restoring force is stored as Elastic Potential Energy.

Derivation:

Elastic Potential Energy ( $U = \frac{1}{2}kx^2$ )

Step 1: Work Done by External Force
To stretch a spring by $dx$ , external force $F_{ext} = kx$ .
$dW = F_{ext} dx = kx \, dx$ .

Step 2: Integrate
Total work to stretch from $0$ to $x$ :
$W = \int_{0}^{x} kx \, dx = k \left[ \dfrac{x^2}{2} \right]_0^x$ .

Step 3: Final Formula
$W = U = \dfrac{1}{2}kx^2$ .

4. Conservation of Mechanical Energy

If only conservative forces act on a system, the Total Mechanical Energy ( $E$ ) remains constant.

E = K + U = \text{Constant}

\Delta K + \Delta U = 0

Motion in a Vertical Circle

For objects like pendulums or roller coasters in vertical loops, mechanical energy conservation applies.

Minimum Speed at Top: $v_{top} = \sqrt{gR}$ (to just complete loop).
Minimum Speed at Bottom: $v_{bottom} = \sqrt{5gR}$ .

Problem 3: A ball is dropped from height $H$ . Find velocity at height $h$ using conservation of energy.

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At top: $K=0, U=mgH$ . Total $E = mgH$ .
At height $h$ : $K = \dfrac{1}{2}mv^2, U = mgh$ .
$mgH = \dfrac{1}{2}mv^2 + mgh$
$v^2 = 2g(H-h) \Rightarrow v = \sqrt{2g(H-h)}$ .

When Non-Conservative Forces Act

If non-conservative forces (like friction, air resistance) do work, mechanical energy is not conserved. The generalized work-energy principle states:

W_{\text{nc}} = \Delta K + \Delta U

where $W_{\text{nc}}$ is the work done by non-conservative forces. This work usually appears as heat or sound.

Problem 4 (Raindrop with Air Resistance): A 1 g raindrop falls from 1 km and hits ground at 50 m/s. Find work done by (a) gravity, (b) air resistance. ( $g = 10$ m/s²)

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(a) $W_g = mgh = 0.001 \times 10 \times 1000 = 10$ J.
(b) $\Delta K = \dfrac{1}{2}mv^2 = 0.5 \times 0.001 \times 2500 = 1.25$ J.
By work-energy theorem: $W_g + W_{\text{air}} = \Delta K$
$10 + W_{\text{air}} = 1.25 \Rightarrow W_{\text{air}} = -8.75$ J.

5. Power & Efficiency

Power is the rate of doing work. It is a scalar quantity.

P_{avg} = \dfrac{W}{t}

P_{inst} = \dfrac{dW}{dt} = \vec{F} \cdot \vec{v}

Unit: Watt (W). 1 Horsepower (hp) = 746 W.

Electricity Bills:

Energy consumption is measured in kilowatt-hours (kWh).
$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$ .
Efficiency ( $\eta$ ) of a device: $\eta = \dfrac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%$ .

Problem 5: An elevator of mass 500 kg is pulled up at a constant speed of 2 m/s. Find the power delivered by the motor. ( $g=10$ m/s²)

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Force required $F = mg = 500 \times 10 = 5000$ N.
Power $P = F v = 5000 \times 2 = 10000$ W = 10 kW.

6. Collisions

A collision is an isolated event where strong forces act for a short time. Momentum is conserved in ALL collisions.

Visual comparison between elastic and inelastic collisions. — Momentum is always conserved; Kinetic Energy is conserved only in Elastic collisions.

Types of Collisions

Type	Momentum	Kinetic Energy
Elastic	Conserved	Conserved
Inelastic	Conserved	Not Conserved (Loss)
Perfectly Inelastic	Conserved	Max Loss (Bodies stick)

Coefficient of Restitution ( $e$ )

The elasticity of a collision is quantified by the coefficient of restitution ( $e$ ):

e = \dfrac{\text{Relative speed after collision}}{\text{Relative speed before collision}} = \dfrac{v_2 - v_1}{u_1 - u_2}

Elastic: $e = 1$
Inelastic: $0 < e < 1$
Perfectly Inelastic: $e = 0$ (bodies stick together)

Problem 6: A ball dropped from 5 m rebounds to 3.2 m. Find $e$ .

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Using $v = \sqrt{2gh}$ ,
$e = \dfrac{\sqrt{2g \cdot 3.2}}{\sqrt{2g \cdot 5}} = \sqrt{\dfrac{3.2}{5}} = \sqrt{0.64} = 0.8$ .

Elastic Collision in 1D

If two bodies of masses $m_1, m_2$ moving with $u_1, u_2$ collide elastically:

Relative Velocity Rule: Velocity of approach = Velocity of separation.
$u_1 - u_2 = v_2 - v_1$

Note on 2D Collisions: In two-dimensional collisions, momentum conservation gives two equations (x and y components). A third equation (e.g., from elasticity or measured angle) is needed to solve for the unknowns.

Problem 7: A 2kg ball moving at 6 m/s collides head-on elastically with a 4kg ball at rest. Find final velocities.

Show Answer

1. Momentum: $2(6) + 0 = 2v_1 + 4v_2 \Rightarrow 12 = 2v_1 + 4v_2 \Rightarrow 6 = v_1 + 2v_2$ .
2. Elasticity: $u_1 - u_2 = v_2 - v_1 \Rightarrow 6 - 0 = v_2 - v_1 \Rightarrow v_1 = v_2 - 6$ .
Substitute (2) in (1): $6 = (v_2 - 6) + 2v_2 \Rightarrow 12 = 3v_2 \Rightarrow v_2 = 4$ m/s.
$v_1 = 4 - 6 = -2$ m/s (Rebounds).

Problem 8 (Perfectly Inelastic): Two 2 kg blocks collide and stick together. Initial speeds were 4 m/s and 2 m/s in the same direction. Find final speed.

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