 ## Work

Subject: Science

#### Overview

Work done by force acting on a body is defined as the product of force and displacement of the body in the direction of the force. This note provide us the information about work and its types.
##### Work

Work done by the force acting on a body is defined as the product of force and displacement of the body in the direction of the force. It is a scalar quantity. Force acting on the body must produce a displacement for the work is to be done by the force. Thus for the work to be done, the following two conditions must be fulfilled:

• A force must be applied, and
• The applied force must produce a displacement in any direction except perpendicular to the direction of the force.

Mathematically,

Work= Force × Displacement (in the direction of the force)

i.e. W= F.d ................(i)

The SI unit of force is newton (N) and that of displacement is a metre (m). So, the unit of work is Newton metre (Nm), which is Joule (J).

Thus, one joule of work is said to be done when one Newton force displaces a body through one metre in its own direction.  Work against friction

Force is to be applied to move, roll, or drag a body over a surface of another body. Example: If an object is dragged to a certain distance (d).

Activity: To demonstrate work against friction.

At first, placed the brick over the surface of the table. Now, pulled the brick from initial position to final position as shown in the figure.It moves by a displacement's'. Here, force is applied against the frictional force between the surface of the brick and the table to set the brick in motion. Here, the work is done against friction.

Work against gravity

Force is applied to lift a body against the force of gravity. Example: If an object is lifted to a certain height (h).

Activity: To demonstrate the work against gravity.

Take a body of mass 'm' and a spring balance to lift the body. Then lift the body of mass 'm' upto the certain height 'h'. The pointer of the spring balance gives the force applied 'F' against the gravity. Here, the work is done against the gravity and it is given by the product of the force 'F' and height 'H'.

#### Measurement of work

On the basis of the direction of motion of a body and the direction of force applied, the way of calculation of work done is determined.

1. When a body moves in the direction of force applied:
In the above conditions (work against friction and work against gravity) the direction of motion of the body and the direction are the same.
Work done (W) = Force (F) × Displacement (d)
2. When a body moves oblique to the direction of force applied:
Consider a body is pulled on an inclined plane AC. The direction of force applied on it is in AC direction but the motion direction is AB.
In This condition,
cosθ = $\frac{Base}{Hypotenuse}$ = $\frac{AB}{AC}$
AB = AC Cosθ
∴W= F Cosθ× D
##### Things to remember
• Work done by force acting on a body is defined as the product of force and displacement of the body in the direction of the force.
• The unit of work is newton-metre (Nm), which is Joule (J).
• One joule of work is said to be done when one Newton force displaces a body through one metre in its own direction.
• When a body moves in the direction of force applied then Work done (W) = Force (F) × Displacement (d).
• When a body moves oblique to the direction of force applied then Work done (W) = Force (F) CosƟ. Distance (D)
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Work

Work is said to be done only when the force applied to the body produces a displacement in the body.
Mathematically,
Work done is by the force acting on a body is defined as the product of force and displacement of the body in the direction of force.
i.e. Work = Force x displacement
W = F x d.

Its SI unit is Joule (J) or Nm.
The amount of work done depends on magnitude and direction of force displacement that it produces
To do work there must be motion of object with the application of force. But here, displacement of the building is zero.
i.e. W = F x 0 = 0
Hence, even if he gets tired, there is no work done.
1 J (1 Joule) work is said to be done if 1 Newton force displaces a body through a distance of 1 meter in the direction of force.
The CGS unit of force is 'erg'.
It is because it does not require any direction for its description.
Yes, it is possible that a force is acting on a body but still the work done is zero. For example, in case of a man pushing a hill.
There are two types of work done:
1. Work done against gravity: While lifting a body to a height, work is done against gravity.
2. Work done against friction: When a body is pushed or pulled then work is done against friction.

When force is applied perpendicular to the displacement, the angle θ = 900.
Hence,
W = F s.cos900 = 0
So, no work is done.

As we know that, W = F x d having unit Nm. So, in terms of fundamental unit it is kgm2/s2.

1 erg work is the work done by a force of 1 dyne to move a body through a distance of 1 cm in the direction of force.
Force (F) = 300N
Distance (d) = 70cm = 0.7m
Work done (W) = ?
By using formula, W = F x d
= 300 x 0. 7
= 210 J
Thus, a man does 210 J work.

Solution:
Here,
Mass (m) = 30kg
Height (h) =1.5km = 1500m
Acceleration due to gravity (g) = 10m/s2
As we know that,
W = mgh
or, W = 30 x 10 x 1500
or, W = 450000J
= 4.5 x 105J

Solution:
Here,
Work done (W) = 50KJ = 50000J
Distance (d) = 10m
Force (F) = ?
By using formula,
W = F x d

or, 50000 = F x 10
or, F = 50000/10
or, F = 5000N

Thus, the force of friction of the ground is 5000N.

Here, Force (F) = 10N
Displacement (d) = 1m
Work done (W) = ?

We have,
W = F.d = 10 $\times$ 1 = 10J
Therefore, work done by the force is 10J.

Here Mass (m1) = 200 kg
Mass of the person (m2) = 50 kg
Height (h) = 10 m
Acceleration due to gravity (g) = $10m/s^{2}$
We have,
W = F.d [ $\therefore$ F = m.g]
= m.g.d [ $\therefore$ m= m1+m2 = 200+50 = 250 kg]
= 250 $\times$ 10 $\times$ 10
= 2.5 $\times$ $10^{4}$J

Thus, work done against the gravity is 2.5 $\times$ $10^{4}$J

Given,
D = 20m
f = 200N
$\theta$ = 15 $^o$

We have,
W = F.d (Cos $\theta$ + Sin $\theta$ )
= 200.20 (Cos 15 $^0$ + Sin 15 $^0$ )
= 200 $\times$ 10 (0.9659 + 0.2588)
= 200 $\times$ 10 $\times$ 1.2247 = 2449.4J

Thus, work done on the slope is 2449.4J