ICSE Class 10 Physics Chapter 02 Work Energy and Power

Work, Energy And Power

Syllabus

Work, energy, power and their relation with force. Definition of work. \(W = FS \cos \theta\); special cases of \(\theta = 0°, 90°\). \(W = mgh\). Definition of energy, energy as work done. Various units of work and energy and their relation with S.I. units. [erg, calorie, kWh and eV]. Definition of power, \(P = W/t\); S.I. and C.G.S. units; other units, kilowatt (kW), megawatt (MW) and gigawatt (GW); and horse power (1 HP = 746 W) [Simple numerical problems on work, power and energy].

Different types of energy (e.g., chemical energy, mechanical energy, heat energy, electrical energy, nuclear energy, sound energy, light energy).

Mechanical energy: potential energy \(U = mgh\) (derivation included), gravitational potential energy, examples; kinetic energy \(K = \frac{1}{2} mv^2\) (derivation included); forms of kinetic energy: translational, rotational and vibrational - only simple examples. [Numerical problems on K and U only in case of translational motion]; qualitative discussions of electrical, chemical, heat, nuclear, light and sound energy, conversion from one form to another; common examples.

Principle of conservation of energy. Statement of the principle of conservation of energy; theoretical verification that \(U + K =\) constant for a freely falling body. Application of this law to simple pendulum (qualitative only); simple numerical problems.

Work, Energy And Power, Their Measurements And Units

Work

In our daily language, the word 'work' is used for some sort of exertion (physical or mental) or for various activities such as while writing, reading or eating, we say that we are doing work. But in Physics, the term 'work' is used in relation to the displacement produced by a force. The work is said to be done only when a body moves under the influence of a force. If there is no displacement of the body even when a force acts on it, the work done is said to be zero. Thus:

Work is said to be done only when the force applied on a body makes the body move (i.e., there is a displacement of the body).

For example, a man while pushing a car (Fig. 2.1), a cyclist while pedalling a cycle, a horse while pulling a cart, a boy going upstairs, a coolie lifting a load, all exert the force which produces motion, so they do work.

However, if a man tries to push a wall (Fig. 2.2) or a child tries to push a heavy stone and they are unable to move it, then scientifically no work is being done by them.

Measurement Of Work

If forces \(F_1\) and \(F_2\) (where \(F_1 > F_2\)) move two different bodies by the same distance, the work done by the force \(F_1\) is said to be more than that by the force \(F_2\). Similarly, if a force \(F\) moves the bodies 1 and 2 by distances \(S_1\) and \(S_2\) respectively (where \(S_1 > S_2\)), the work done by the force \(F\) on body 1 is said to be more than on body 2. Thus the amount of work done depends on both factors: the magnitude of the force applied, and the magnitude of the displacement.

The amount of work done by a force is equal to the product of the force and the displacement of the point of application of the force in the direction of force.

i.e., Work = Force × displacement of the point of application of force in the direction of force

\[W = F \times S\]

Here we assume that the force does not change during the displacement and it acts throughout the displacement.

In eqn. (2.1), if \(S = 0\), then \(W = 0\). Thus:

If a force acts on a body and the body does not move i.e., displacement is zero, then no work is done.

Work is a scalar quantity.

Expression Of Work (\(W = FS \cos \theta\))

It is not necessary that the force always causes the displacement of the body in its own direction. If under some circumstances, a force displaces the body in a direction other than the direction of force, then we can determine the amount of work done by the force in two ways: (1) by finding the component of displacement of the body in the direction of force, and (2) by finding the component of force in the direction of displacement.

(1) By finding the component of displacement along the force: In Fig. 2.4, suppose a constant force F acts on a body along AB and displaces the body on an inclined surface from A to C. The displacement of the body is \(AC = S\), which is at an angle \(\theta\) to the direction of force. To find the component of displacement in the direction of force (i.e., along AB), a perpendicular CB is drawn from the point C on AB. Then the component of displacement in the direction of force is AB.

Hence, work done \(W = F \times AB\)

But in the right angled triangle ABC,

\[\cos \theta = \frac{\text{base}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{AB}{S}\]

or \(AB = S \cos \theta\)

Hence, \[W = F \times S \cos \theta\]

or W = Force × component of displacement in the direction of force

(2) By finding the component of force along the displacement: In Fig. 2.5, if PA represents the magnitude and direction of force F acting on the body, then the component of force F in the direction of displacement (i.e., along AC) is NA.

Therefore, work done

\[W = NA \times AC = NA \times S\]

But from right-angled triangle PNA,

\(NA = PA \cos \theta = F \cos \theta\)

Then work done \[W = F \cos \theta \times S\]

or Work = Component of force in the direction of displacement × displacement

From eqns. (2.2) or (2.3), it is clear that the work done is equal to the product of (i) magnitude of force F, (ii) magnitude of displacement S, and (iii) cosine of the angle \(\theta\) between the directions of force F and displacement S (i.e., \(\cos \theta\)).

Note: Since force F and displacement S are vector quantities and work W is a scalar quantity, so work is expressed as the dot product (or scalar product) of force and displacement vectors. The dot product of two vectors is a scalar. In vector form, work done W is written as \[W = \vec{F} \cdot \vec{S}\]

Special Cases

Case (i): If the displacement is in the direction of force, i.e., \(\theta = 0°\), then \(\cos 0° = 1\).

\(\therefore W = F \times S\)

The work done is positive.

Examples: (1) In free fall of a body of mass m under gravity through a height h from A to B (Fig. 2.6), the force of gravity \(F (= mg)\) is in the direction of displacement \(S (= h)\) and the work done by the force of gravity is \(W = FS = mgh\).

(2) A coolie does work on the load when he raises it up against the force of gravity. Both the force exerted by coolie \(F (= mg)\) and the displacement \((= h)\), are in upward direction. The work done in raising the load = mgh.

Case (ii): If the displacement is normal to the direction of force, i.e., \(\theta = 90°\), then \(\cos 90° = 0\).

\(\therefore W = 0\)

Hence the work done is zero.

Examples: (1) When a coolie walks on a horizontal ground while carrying a load on his head, no work is done against the force of gravity because the displacement of load is normal to the direction of force of gravity (which is vertically downwards).

(2) When a body moves in a circular path in a horizontal plane, no work is done since the centripetal force on the body is directed towards the centre of circular path and the displacement at all instants is along the tangent to the circular path, i.e., normal to the direction of force on the body as shown in Fig. 2.7. It is for this reason that in a circular path, the kinetic energy and hence the speed of the body does not change although a force acts on the body. Further on completion of one complete rotation, the total displacement of the body becomes zero, so the work done in one complete rotation is zero.

Conditions For The Work Done By A Force To Be Zero: From the above discussion, it is clear that the amount of work done by a force is zero in the following two situations:

(1) when there is no displacement (\(S = 0\)), and

(2) when the displacement is normal to the direction of force (\(\theta = 90°\)).

Case (iii): If the displacement is in a direction opposite to the force, i.e., \(\theta = 180°\), then \(\cos 180° = -1\).

\(\therefore W = - F \times S\)

The work done is negative. This is usually the case when the force opposes the motion or it tries to stop a moving body.

Examples: (1) When a body moves on a surface, the force of friction between the body and surface is in direction opposite to the motion of the body, therefore, the work done by the force of friction is negative.

(2) When a ball of mass m is thrown upwards from A to B to a height h (Fig. 2.8), the displacement h (upwards) is opposite to the direction of force of gravity mg (downwards), so the work done by the force of gravity mg in displacement h is \(W = -mgh\) i.e., negative.

Note: If the force is variable (i.e., force varies during the displacement), the work done is determined by plotting a force-displacement graph. The force is taken on Y-axis and the displacement (in the direction of force) is taken on X-axis. The area enclosed by the sketch and the displacement axis (i.e., X-axis) gives the work done. In Fig. 2.9, the force F is directly proportional to the displacement and the graph for force against displacement is an inclined straight line OA. The work done by the force in displacement S is equal to the area of the triangle OAB (\(= \frac{1}{2} F \times S\)) which is shown shaded in Fig. 2.9.

This method is applicable in all situations. If the force is constant (i.e., it does not change with the displacement), the sketch on graph will be a straight line parallel to the X-axis and the area of the rectangle enclosed between the straight line and the X-axis will be equal to the work done.

Work Done By The Force Of Gravity \(W = mgh\)

Let a body of mass m be moved down through a vertical height h either directly or through an inclined plane (e.g. a hill, slope or stairs). The force of gravity on the body is \(F = mg\) acting vertically downwards and the vertical displacement in the direction of force is \(S = h\). Therefore the work done by the force of gravity is

\[W = FS = mgh\]

Thus work done by the force of gravity is same whether a person comes down from a certain height using the stairs or slope or a lift (or elevator).

Similarly, if a boy of mass m goes up through a vertical height h either directly or through the stairs or slope or lift, the work \(W = - mgh\) is done by the force of gravity on the boy (or the work \(W = mgh\) is done by the boy against the force of gravity).

Units Of Work

S.I. unit: The S.I. unit of work is joule. It is abbreviated as J.

Since work = force × displacement

\(\therefore\) 1 joule = 1 newton × 1 metre

Thus,

1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.

Bigger units of work are kilo-joule (kJ), mega-joule (MJ) and giga-joule (GJ), where 1 kJ = 10³ J, 1 MJ = 10⁶ J and 1 GJ = 10⁹ J.

C.G.S. unit: The C.G.S. unit of work is erg, where 1 erg = 1 dyne × 1 cm.

Thus,

1 erg of work is said to be done when a force of 1 dyne displaces a body through a distance of 1 cm in its own direction.

Relationship Between Joule And Erg

1 joule = 1 N × 1 m

But 1 N = 10⁵ dyne and 1 m = 10² cm

\(\therefore\) 1 joule = 10⁵ dyne × 10² cm = 10⁷ dyne × cm = 10⁷ erg

Thus \[1 \text{ joule} = 10^7 \text{ erg}\]

Teacher's Note

Work in physics requires both force and displacement in the same direction - this is why pushing a wall does no work, even though you feel tired. Understanding this distinction helps explain why climbing stairs requires effort but walking horizontally at constant speed does not increase your mechanical energy.

This is a preview of the first 3 pages. To get the complete book, click below.