"Work, Energy and Power"

About

Work, energy, and power are among the most fundamental concepts in physics. This chapter connects force and motion through the idea of work, introduces kinetic and potential energy, and explains one of nature's most important principles — the conservation of mechanical energy. You will also learn to calculate power and analyse collisions.

Key Concepts

6.1 Work

Work is said to be done when a force acts on a body and the body moves in the direction of the force (or has a component of displacement along the force).

$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$

Where $\theta$ is the angle between the force vector and displacement vector.

Work is a scalar quantity. SI unit: joule (J) = N⋅m

Sign of work:

Work done by centripetal force is always zero because the force is perpendicular to the displacement at every instant.

Work done by gravity when lifting a mass $m$ through height $h$: $W_g=-mgh$ (negative because gravity opposes the upward displacement).

6.2 Work Done by a Variable Force

When force varies with position, work is the area under the force-displacement graph:

$W={\int }_{x_i}^{x_f}F(x)dx$

Graphically: Area under F-x curve = Work done.

6.3 Energy

Energy is the capacity to do work. SI unit: joule (J).

Kinetic Energy (KE)

Energy possessed by a body due to its motion:

$K=\frac{1}{2}m{v}^2$

• Always positive
• Depends on mass and speed
• Frame-dependent

Potential Energy (PE)

Energy possessed by a body due to its position or configuration.

Gravitational PE: $U=mgh$ (near Earth's surface)

Elastic PE (spring): $U=\frac{1}{2}k{x}^2$

6.4 Work-Energy Theorem

The net work done on a body equals the change in its kinetic energy:

$W_{\text{net}}=\Delta K=K_f-K_i=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$

6.5 Conservation of Mechanical Energy

When only conservative forces (gravity, spring force) act on a system, the total mechanical energy remains constant:

$K_i+U_i=K_f+U_f$

$\frac{1}{2}m{v}_i^2+mg{h}_i=\frac{1}{2}m{v}_f^2+mg{h}_f$

Conservative forces: Work done is path-independent (gravity, electrostatic, spring force).

Non-conservative forces: Work done depends on the path (friction, air resistance, viscous drag).

6.6 Power

Power is the rate of doing work or the rate of transfer of energy.

$P=\frac{W}{t}$

Instantaneous power: $P=\vec{F}\cdot \vec{v}$

SI unit: watt (W) = 1 J/s

Larger units: 1 kW = 10³ W, 1 MW = 10⁶ W, 1 HP = 746 W

6.7 Collisions

Elastic collision: Both momentum AND kinetic energy are conserved.

Inelastic collision: Only momentum is conserved; kinetic energy is NOT conserved (converted to heat, sound, deformation).

Perfectly inelastic collision: Bodies stick together after collision.

INTEXT QUESTIONS 6.1

Q1. When a particle rotates in a circle, a force acts on the particle. Calculate the work done by this force on the particle.

Ans: When a particle moves in a circular path, the centripetal force always acts toward the centre of the circle. At any instant, the displacement of the particle is along the tangent to the circle. Since the centripetal force is radial (toward centre) and displacement is tangential, the angle between force and displacement is always 90°.

$W=Fd\cos 90°=Fd\times 0=0$

The work done by the centripetal force is zero.

Q2. Give one example of each of the following. Work done by a force is: (a) zero, (b) negative, (c) positive.

Ans:

(a) Zero work: Carrying a bag while walking horizontally on level ground — the applied force (upward) is perpendicular to the displacement (horizontal).

(b) Negative work: Applying brakes to stop a moving car — friction opposes the motion ($\theta =180°$).

(c) Positive work: Pushing a box along the floor in the direction of the applied force — force and displacement are in the same direction ($\theta =0°$).

Q3. A bag of grains of mass 2 kg is lifted through a height of 5 m.

(a) How much work is done by the lift force?

Ans:

• $m=2$ kg, $h=5$ m, $g=9.8$ m/s²
• Lift force = $mg=2\times 9.8=19.6$ N (upward)
• Displacement = 5 m (upward), $\theta =0°$

$W=Fd\cos 0°=19.6\times 5\times 1=98\text{ J}$

(b) How much work is done by the force of gravity?

Ans:

• Gravitational force = $mg=19.6$ N (downward)
• Displacement = 5 m (upward), $\theta =180°$

$W=Fd\cos 180°=19.6\times 5\times (-1)=-98\text{ J}$

Terminal Exercise

1. Define work. Under what conditions is the work done by a force (a) positive, (b) negative, (c) zero? Give one example of each.

2. State and prove the work-energy theorem for a constant force.

3. A body of mass 5 kg is moving with a velocity of 10 m/s. A force is applied to it so that in 25 seconds, it attains a velocity of 35 m/s. Calculate the work done.

4. Define kinetic energy. Derive the expression $K=\frac{1}{2}m{v}^2$.

5. Define potential energy. Derive the expression for gravitational potential energy: $U=mgh$.

6. State the law of conservation of mechanical energy. A body of mass 2 kg is dropped from a height of 20 m. Using energy conservation, find its velocity just before hitting the ground. (g = 10 m/s²)

7. Show that the total mechanical energy of a freely falling body remains constant.

8. Define power. Show that $P=\vec{F}\cdot \vec{v}$. An engine pumps 1000 kg of water per minute to a height of 20 m. Find the power of the engine. (g = 10 m/s²)

9. Distinguish between elastic and inelastic collisions. Give two examples of each.

10. A ball of mass 0.1 kg moving with a velocity of 10 m/s strikes a wall and rebounds with the same speed. Find the change in momentum and the impulse imparted. If the contact time is 0.01 s, find the average force exerted by the wall.

11. A force $\vec{F}=(3\hat{i}+4\hat{j})$ N acts on a particle and displaces it by $\vec{d}=(2\hat{i}-\hat{j})$ m. Calculate the work done.

12. The kinetic energy of a body of mass 2 kg is 100 J. What is its momentum?

Worked Examples

Example 1: Work by a Constant Force

Problem: A force of 50 N pulls a box through a distance of 10 m at an angle of 60° with the horizontal. Calculate the work done.

Solution: $W=Fd\cos \theta =50\times 10\times \cos 60°=500\times 0.5=250\text{ J}$

Example 2: Work-Energy Theorem

Problem: A car of mass 1000 kg moving at 20 m/s is brought to rest by applying brakes. Find the work done by the braking force.

Solution: By work-energy theorem: $W=\Delta K=\frac{1}{2}m(v_f^2-v_i^2)=\frac{1}{2}\times 1000\times (0-400)=-200,000\text{ J}$

The negative sign indicates work done against the motion.

Example 3: Conservation of Energy

Problem: A stone of mass 0.5 kg is thrown vertically upward with a speed of 20 m/s. Find the maximum height reached. (g = 10 m/s²)

Solution: By conservation of energy: $\frac{1}{2}m{v}_i^2=mg{h}_{\text{max}}$

$h_{\text{max}}=\frac{v_i^2}{2g}=\frac{400}{20}=20\text{ m}$

Example 4: Power

Problem: A pump lifts 200 kg of water through a height of 10 m in 20 seconds. Find the power. (g = 10 m/s²)

Solution: Work done = $mgh=200\times 10\times 10=20,000$ J

$P=\frac{W}{t}=\frac{20,000}{20}=1000\text{ W}=1\text{ kW}$

Common Mistakes

1. Forgetting $\cos \theta$ in work formula: Only the component of force along displacement does work.
2. Thinking centripetal force does work: It doesn't — force is always perpendicular to displacement in uniform circular motion.
3. Confusing power with energy: Power is the RATE of doing work (J/s), not the total work.
4. Assuming kinetic energy is always conserved in collisions: Only in perfectly elastic collisions.
5. Using $U=mgh$ for large heights: Valid only near Earth's surface where g is approximately constant.

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