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Concept of Work

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  • Introduction of Work
  • Work Done at Different Angles and Its Characteristics

Introduction of Work:

“The work done by the force is defined to be the product of the component of the force in the direction of the displacement and the magnitude of this displacement.”

The work W done by a constant force F when its point of application undergoes a displacement s is defined to be
`W=bar F.bar s=F s cos θ`

where θ is the angle between `bar F` and `bar s` as indicated in the figure.  Only the component of `bar F` along s, that is, fcos θ, contributes to the work done.
Work is a scalar quantity and its SI unit is the joule (J). From the above equation,  we see that
1 J = 1 Nm = 107 erg
In terms of rectangular components, the two vectors are
`bar F= F_x hat i + F_y hat j + F_ z hat k`
and `bar s = Δx hat i +Δ y hat j+Δ z hat k`
hence, the above equation may be written as
`W=F_x Δx + F_y Δy + F_z Δz`

The work done by a given force on a body depends only on the force, the displacement, and the angle between them. It does not depend on the velocity or the acceleration of the body or on the presence of other forces.

When several forces act on a body, one may calculate the work done by each force individually. The net work done on the body is the algebraic sum of individual contributions.
`W_"net" = bar F_1. bar S_1 + bar F_2. bar S_2 +.........+ bar F_n. bar S_n` or
`W_"net" = W_1 + W_2 +............ + W_n`

Let us consider some situations:

  • Push a pebble lying on a surface. The pebble moves through a distance. You exerted force on the pebble and the pebble got displaced. In this situation, work is done.
  • A girl pulls a trolley and the trolley moves through a distance. The girl has exerted a force on the trolley and it is displaced. Therefore, work is done.
  • Lift a book through a height. To do this, you must apply force. The book rises up. A force acts on the book, causing it to move. Therefore, the task is complete.

A closer look at the above situations reveals that two conditions need to be satisfied for work to be done:

  • A force should act on an object.
  • The object must be displaced.

No work is done if:

1. The displacement is zero

2. The force is zero.

3. The force and displacement are mutually perpendicular.

Work Done at Different Angles and Its Characteristics:

 Displacement of an object

In the scenarios shown in images B and C, when a child pulls a toy with a string or a large vehicle tows a smaller one, the direction of the applied force differs from the direction of displacement. In such cases, the force is applied at an angle to the direction of movement. To calculate the work done in these situations, the applied force needs to be resolved into a component that acts along the direction of displacement.

Let represent the applied force and F1 its component in the direction of displacement. If the displacement is , the work done is calculated as:

W=F1⋅s

Here, the force is applied at an angle θ with respect to the horizontal direction. Using trigonometry, the horizontal component of the force (F1) can be expressed as:

F₁ = Fcos⁡θ

Thus, the work done can be written as:

W=Fcos⁡θ⋅s

This method allows calculating work when the force and displacement are not aligned but form an angle.

Force used for the displacement 

cos θ = base / hypotenuse

cos θ = `(F_1)/(F)`

F₁ = Fcos θ

Thus, the work done by F1 is

W = F cos θ s

W = F s cos θ

Conclusions about the work done for the specific values of θ in the following table.

θ cos θ W = F s cos θ Conclusion
1 W = F s Work is maximum and positive.
90° 0 0 No work is done.
180° -1 W = -F s Work is negative.

Unit of work:

Work = Force × Displacement

  • In the SI system, the unit of force is Newton (N) and the unit of displacement is metre (m). Thus, the unit of force is the newton-meter. This is called a joule.
  • 1 Joule : If a force of 1 Newton displaces an object through 1 metre in the direction of the force, the amount of work done on the object is 1 joule. 1 joule = 1 newton × 1 metre
    1 J = 1 N × 1 m
  • In the CGS system, the unit of force is dyne and that of displacement is centimetre (cm). Thus, the unit of work done is dyne-centimetre. This is called an erg.
  • 1 erg: If a force of 1 dyne displaces an object through 1 centimetre in the direction of the force, the amount of work done is 1 erg. 1 erg = 1 dyne × 1 cm

Relationship between joule and erg:

We know that 1 newton = 105 dyne and 1 m = 102 cm
Work = force × displacement
1 joule = 1 newton × 1 m
1 joule = 105 dyne× 102 cm = 107 dyne cm
1 joule = 107 erg

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