Class 11 Physics · NCERT notes for NEET

Work, Energy and Power: NCERT Notes, Formulas and Tricks for NEET

How forces transfer energy: work, kinetic and potential energy, conservation of energy, power, collisions and motion in a vertical circle. NEET asks from this chapter regularly, and its energy method is used again in rotation, gravitation, SHM and electrostatics.

1. Work done by constant and variable forces
2. Kinetic and potential energy, work-energy theorem 🔒
3. Conservation of mechanical energy 🔒
4. Power 🔒
5. Collisions 🔒
6. Motion in a vertical circle 🔒

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Topic 1

Work done by constant and variable forces

Carry a heavy school bag along a flat road for an hour and you feel tired, yet the work you do on the bag is almost zero: your force is up, the bag moves sideways.

Scalar (dot) product: $A \cdot B = A B cos θ$ . It is commutative and distributive, and it is zero when the vectors are perpendicular.
Unit vectors: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$ , so $A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ .
Work by a constant force = (component of force along the displacement) $\times$ displacement: $W = F \cdot d = F d cos θ$ .
Work is a scalar. SI unit joule, $1 J = 1 N m$ ; dimensions $[M L^{2} T^{- 2}]$ .
No work is done if the displacement is zero, the force is zero, or the force is perpendicular to the displacement ( $θ = 9 0^{\circ}$ ).
Work is positive for $0^{\circ} \leq θ < 9 0^{\circ}$ and negative for $9 0^{\circ} < θ \leq 18 0^{\circ}$ . Friction on a sliding block and gravity on a rising ball do negative work.
Variable force: $W = \int_{x_{i}}^{x_{f}} F (x) d x$ = area under the $F$ - $x$ graph. Area below the $x$ -axis counts as negative.
Work depends on the frame of reference, because displacement does.
Other energy units: $1 erg = 1 0^{- 7} J$ , $1 eV = 1.6 \times 1 0^{- 19} J$ , $1 cal = 4.186 J$ , $1 kWh = 3.6 \times 1 0^{6} J$ .

Only the part of F along the displacement (Fcostheta) does work. The sign of costheta decides the sign of work: positive, zero or negative. — Only the part of $F$ along the displacement ( $F cos θ$ ) does work. The sign of $cos θ$ decides the sign of work: positive, zero or negative.

For a variable force, work is the area under the F-x graph. Area above the axis is positive, area below is negative: here 28-7=21 J. — For a variable force, work is the area under the $F$ - $x$ graph. Area above the axis is positive, area below is negative: here $28 - 7 = 21$ J.

Formulas

$W = F \cdot d = F d cos θ$
Constant force; $θ$ = angle between $F$ and $d$
$W = F_{x} d_{x} + F_{y} d_{y} + F_{z} d_{z}$
When vectors are given in $\hat{i}, \hat{j}, \hat{k}$ form
$W = \int_{x_{i}}^{x_{f}} F (x) d x$
Variable force; equals signed area under the $F$ - $x$ graph
$cos θ = \frac{A \cdot B}{A B}$
Angle between two vectors

Tips, tricks and hacks

Dot product in 5 secondsMultiply matching components and add.

F = (3 \hat{i} + 4 \hat{j})

d = (2 \hat{i} - \hat{j})

W = 6 - 4 = 2

J. No need to find angles.

Perpendicular means zeroSpot forces that are always perpendicular to motion: centripetal force, normal force on a level road, tension in a uniformly whirled string. Their work is 0 without any calculation.

Gravity: only height mattersWork by gravity

= m g \times

(vertical drop), whatever the path. A 2 kg stone sliding down any 5 m high slope gets

2 \times 10 \times 5 = 100

J from gravity.

Areas, not integralsFor a graph made of straight lines, split it into triangles and rectangles and add with signs. It is faster than writing the equation of each line.

Common mistakes

✗ Holding a heavy load still (or walking on level ground with it) is hard work, so $W > 0$ .

✓ Displacement along the force is zero, so physical work on the load is zero; your muscles spend energy internally.

✗ Work is a vector because force and displacement are vectors.

✓ The dot product of two vectors is a scalar; work has sign but no direction.

✗ Area under the $F$ - $x$ curve below the axis is added as positive.

✓ Area below the axis means force opposite to displacement, so that work is negative.

NEET traps

NEET asks for the work done by a particular force (for example friction or gravity), not the net force. Only the net-force work equals the change in KE.

Static friction can do positive work on a body in the ground frame: a box on an accelerating truck is pushed forward by static friction.

A force given as $F = (\dots)$ with displacement between two points: use $d = r_{2} - r_{1}$ , not $r_{2}$ alone.

Quick check: A force

F = (3 \hat{i} + 4 \hat{j})

N moves a particle from

(1, 2)

m to

(3, 1)

m. The work done is

2 J
10 J
5 J
0 J

Answer: A. $d = (2 \hat{i} - \hat{j})$ m, so $W = 3 (2) + 4 (- 1) = 2$ J.

Topic 2 · 🔒