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Work, Energy & Power

ICSE Class 10 Physics • Chapter 02 Created by Team Vardaan with ❤️ Powered by vardaan comet
Topics to be covered

(A) WORK, POWER AND ENERGY

2.1 WORK

In Physics, the term 'work' is used only when a force produces displacement in a body.

Definition: Work is said to be done only when the force applied on a body makes the body move.

Fig 2.1

Fig 2.1 A man pushing a car

If there is no displacement (e.g., pushing a rigid wall or holding a heavy load while standing still), the work done is zero.

2.2 MEASUREMENT OF WORK

GENERAL FORMULA
$$W = F \times S \cos \theta$$

Where $F$ is force, $S$ is displacement, and $\theta$ is the angle between them.

Fig 2.2

Special Cases ($\theta$):

(i) $\theta = 0^\circ$ (Positive Work): Force & displacement in same direction. $W = F \times S$.

Fig 2.3

(ii) $\theta = 90^\circ$ (Zero Work): Force & displacement are perpendicular. $W = 0$ (e.g., Circular orbit).

Fig 2.4

(iii) $\theta = 180^\circ$ (Negative Work): Displacement opposite to force. $W = -F \times S$ (e.g., Friction).

Fig 2.5

Work Done by a Variable Force

Force-Displacement Graph

The area under a Force-Displacement graph gives the work done.

2.3 WORK DONE BY FORCE OF GRAVITY

GRAVITATIONAL WORK
$$W = mgh$$

where $h$ is the vertical height climbed or descended.

Fig 2.6
ZERO WORK ALERT If a coolie moves horizontally, work done by gravity is ZERO ($\theta = 90^\circ$). Work is only done when lifting the load ($mgh$).

2.4 UNITS OF WORK

RELATIONSHIP J vs ERG
$$1 \text{ joule} = 10^7 \text{ erg}$$

2.5 POWER ($P = W/t$)

Power is the rate of doing work.

POWER FORMULAS
$$P = \frac{W}{t} \quad \text{or} \quad P = F \times v$$

Units of Power:

Difference between Work and Power

Work Power
Force $\times$ Displacement. Rate of doing work.
Independent of time. Depends on time.
NUMERICAL Power Calculation (Crane)

A crane pulls up a car of mass $500 \text{ kg}$ to a vertical height of $4 \text{ m}$ in $20 \text{ s}$. Calculate power. ($g = 9.8$)

Solution:
$W = F \times h = (500 \times 9.8) \times 4 = 19600 \text{ J}$.
$P = W/t = 19600 / 20 = 980 \text{ W}$.

2.7 & 2.8 ENERGY AND ITS UNITS

Definition: Energy is the capacity to do work. It is a scalar quantity with the same units as work (Joules/Ergs).

COMMERCIAL & OTHER UNITS

Energy Transformations

In various devices, energy is converted from one form to another. Examples:

DEGRADED ENERGY During conversion, part of the energy is lost as non-useful forms (like heat due to friction). This is called dissipation of energy or degraded energy.

(B) DIFFERENT FORMS OF ENERGY

2.9 - 2.11 POTENTIAL ENERGY (U)

The energy possessed by a body due to its changed position or configuration.

2.12 KINETIC ENERGY (K)

The energy possessed by a body due to its state of motion.

KINETIC ENERGY & MOMENTUM
$$K = \frac{1}{2} mv^2 \quad | \quad p = \sqrt{2mK}$$
Fig 2.9
BOARD FAVORITE If two bodies (light and heavy) have the same momentum, the lighter body will have more Kinetic Energy (since $K \propto 1/m$).

2.13 WORK-ENERGY THEOREM

Work done by a force on a moving body equals the increase in its kinetic energy: $W = K_f - K_i = \Delta K$.

Fig 2.11

(C) CONSERVATION OF ENERGY

2.16 PRINCIPLE OF CONSERVATION OF ENERGY

Definition: Energy can neither be created nor destroyed; it only changes form. Total mechanical energy ($K + U$) remains constant in absence of friction.

2.17 THEORETICAL VERIFICATION (FREE FALL)

For a freely falling body, total energy $E = mgh$ at all points.

Fig 2.15
Position Kinetic (K) Potential (U) Total (E)
A (Highest) 0 $mgh$ $mgh$
B (Middle) $\frac{1}{2}mgh$ $\frac{1}{2}mgh$ $mgh$
C (Ground) $mgh$ 0 $mgh$

2.18 SIMPLE PENDULUM

Fig 2.17

MASTER NUMERICAL SET (Top 5 Essential Qs)

Q1 Basic Power & Work

A boy of mass $40 \text{ kg}$ climbs $8 \text{ m}$ in $5 \text{ s}$. Power? ($g=10$)

Ans: $P = \frac{mgh}{t} = \frac{40 \times 10 \times 8}{5} = 640 \text{ W}$.

Q2 Efficiency (Hydro Dam)

In a dam, water falls at $1000 \text{ kg/s}$ from $100 \text{ m}$. Calculate power if efficiency is $60\%$.

Ans: 1s Potential Energy = $mgh = 1000 \times 10 \times 100 = 10^6 \text{ J}$.
Power Generated = $60\% \text{ of } 10^6 = 600,000 \text{ W} = 600 \text{ kW}$.

Q3 Inclined Plane Logic Fig 2.12

A block of mass $30 \text{ kg}$ is pulled up a slope of length $3 \text{ m}$ and height $1.5 \text{ m}$ with force $200 \text{ N}$. Calculate: (i) Work done, (ii) Potential energy gain.

Ans: (i) $W = F \times \text{length} = 200 \times 3 = 600 \text{ J}$.
(ii) $U = mgh = 30 \times 10 \times 1.5 = 450 \text{ J}$. (The $150 \text{ J}$ difference is lost to friction).

Q4 Rebound & Energy Loss

A $10\text{ g}$ ball falls from $5\text{ m}$ and rebounds to $4\text{ m}$. Find loss in energy. ($g=9.8$)

Ans: Initial PE = $0.01 \times 9.8 \times 5 = 0.49\text{ J}$.
Final PE = $0.01 \times 9.8 \times 4 = 0.392\text{ J}$.
$\text{Loss} = 0.49 - 0.392 = 0.098\text{ J}$ (Lost as heat/sound).

Q5 Ski Jump Conservation Fig 2.18

A skier ($60 \text{ kg}$) takes off from A ($75 \text{ m}$) to B ($15 \text{ m}$). Calculate speed at B if $75\%$ energy becomes KE.

Ans: $\Delta U = mg(h_1 - h_2) = 60 \times 10 \times (60) = 36000 \text{ J}$.
$K = 0.75 \times 36000 = 27000 \text{ J}$.
$\frac{1}{2}mv^2 = 27000 \implies \frac{1}{2}(60)v^2 = 27000 \implies v^2 = 900 \implies v = 30 \text{ m/s}$.

End of Chapter 02

Practice makes perfect. Review the numericals thrice.

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