Work and Energy — NCERT Solutions, AI Tutor & Practice

Scientific definition of work: Work is said to be done by a force on an object when the force produces a displacement of the object in the direction of the force.

Two conditions must be satisfied for work to be done:

1. A force must act on the object.
2. The object must be displaced (move) as a result of this force.

The work done is given by: $W = F \times s$ where $F$ is the force applied and $s$ is the displacement in the direction of the force.

Unit: The SI unit of work is the joule (J). One joule is the work done when a force of $1\ \text{N}$ moves an object through a distance of $1\ \text{m}$ in the direction of the force. $1\ \text{J} = 1\ \text{N} \times 1\ \text{m}$

Given: Force $F = 15\ \text{N}$, Displacement $s = 4\ \text{m}$.

Formula: $W = F \times s$

Substitution: $W = 15\ \text{N} \times 4\ \text{m}$ $W = 60\ \text{J}$

The work done on the object is 60 joules.

Understand: Work is done only when a force causes a displacement in the direction of the force.

Analyse: While holding the suitcase stationary, the porter applies an upward force equal to the weight of the suitcase: $F = mg = 12\ \text{kg} \times 10\ \text{m/s}^2 = 120\ \text{N}$ However, the suitcase does not move, so the displacement is: $s = 0$

Conclude: Work done, $W = F \times s = 120\ \text{N} \times 0 = 0\ \text{J}$

Hence, no work is done on the suitcase while it is merely held stationary, even though the porter feels tired.

Kinetic energy: The energy possessed by a body due to its motion is called its kinetic energy.

Derivation: Consider an object of mass $m$ at rest. Let a constant force $F$ act on it, giving it acceleration $a$ and moving it a distance $s$, until it reaches velocity $v$.

The work done by the force is: $W = F \times s$

Using Newton's second law, $F = ma$.

Using the equation of motion $v^2 = u^2 + 2as$ with initial velocity $u = 0$: $v^2 = 2as \implies s = \frac{v^2}{2a}$

Substituting into the work equation: $W = ma \times \frac{v^2}{2a} = \frac{1}{2}mv^2$

This work done gets stored as the kinetic energy of the object. Therefore: $E_k = \frac{1}{2}mv^2$

Given: Mass $m = 500\ \text{g} = 0.5\ \text{kg}$, Velocity $v = 10\ \text{m/s}$.

Formula: $E_k = \frac{1}{2}mv^2$

Substitution: $E_k = \frac{1}{2} \times 0.5 \times (10)^2$ $E_k = \frac{1}{2} \times 0.5 \times 100$ $E_k = 25\ \text{J}$

The kinetic energy of the ball is 25 joules.

Given: Mass $m = 8\ \text{kg}$, Height $h = 5\ \text{m}$, $g = 9.8\ \text{m/s}^2$.

Formula for gravitational potential energy: $E_p = mgh$

Substitution: $E_p = 8 \times 9.8 \times 5$ $E_p = 392\ \text{J}$

The potential energy gained by the object is 392 joules.

Law of conservation of energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.

Applying to the falling object:

At the top (height 20 m), the object is at rest, so all its energy is potential energy: $E_p = mgh = 2 \times 10 \times 20 = 400\ \text{J}$

Just before hitting the ground, all potential energy is converted into kinetic energy (ignoring air resistance): $E_k = E_p = 400\ \text{J}$

Hence, the kinetic energy just before hitting the ground is 400 joules.

Power: Power is the rate of doing work, or the rate at which energy is transferred. $P = \frac{W}{t}$

SI unit: The SI unit of power is the watt (W). One watt is the power of a device that does work at the rate of one joule per second. $1\ \text{W} = 1\ \text{J/s}$

Calculation: Given: Work $W = 6000\ \text{J}$, Time $t = 20\ \text{s}$. $P = \frac{W}{t} = \frac{6000}{20} = 300\ \text{W}$

The power of the machine is 300 watts.

Understand: When a body moves with uniform velocity, its acceleration is zero.

Analyse: By Newton's second law, if acceleration $a = 0$, then the net force on the body is: $F_{net} = ma = m \times 0 = 0$

Since work done by the net force is: $W = F_{net} \times s = 0 \times s = 0$

Conclude: No net work is done on a body moving with uniform velocity, because the net force acting on it is zero. (Individual forces such as friction and applied force may do work, but they cancel out.)

Given: Mass $m = 40\ \text{kg}$, Height $h = 4\ \text{m}$, Time $t = 8\ \text{s}$, $g = 10\ \text{m/s}^2$.

Step 1 — Work done against gravity: $W = mgh = 40 \times 10 \times 4 = 1600\ \text{J}$

Step 2 — Power developed: $P = \frac{W}{t} = \frac{1600}{8} = 200\ \text{W}$

The power developed by the boy is 200 watts.

1 kilowatt hour (kWh): It is the commercial unit of energy. One kilowatt hour is the energy consumed by a device of power 1 kilowatt (1000 W) working for 1 hour.

Conversion to joules: $1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s}$ $1\ \text{kWh} = 3\,600\,000\ \text{J} = 3.6 \times 10^6\ \text{J}$

Calculation: Given: Power $P = 1500\ \text{W} = 1.5\ \text{kW}$, Time $t = 4\ \text{h}$. $E = P \times t = 1.5\ \text{kW} \times 4\ \text{h} = 6\ \text{kWh}$

The energy consumed is 6 kWh (also called 6 units).

Given: Mass $m = 5\ \text{kg}$, Total height $h = 10\ \text{m}$, $g = 10\ \text{m/s}^2$.

At the top (h = 10 m, at rest): $E_p = mgh = 5 \times 10 \times 10 = 500\ \text{J}, \quad E_k = 0$ $\text{Total} = 500 + 0 = 500\ \text{J}$

At the midpoint (h = 5 m): $E_p = 5 \times 10 \times 5 = 250\ \text{J}$ Velocity at 5 m fallen: $v^2 = 2g \times 5 = 2 \times 10 \times 5 = 100$ $E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \times 100 = 250\ \text{J}$ $\text{Total} = 250 + 250 = 500\ \text{J}$

Just before hitting the ground (h = 0): $E_p = 0$ Velocity: $v^2 = 2g \times 10 = 2 \times 10 \times 10 = 200$ $E_k = \frac{1}{2} \times 5 \times 200 = 500\ \text{J}$ $\text{Total} = 0 + 500 = 500\ \text{J}$

Conclusion: At all three points the total mechanical energy is 500 J, verifying the law of conservation of energy.