Work and Energy Class 9

 📝 Work and Energy 

SCIENTIFIC CONCEPTION OF WORK

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.

■ Two conditions need to be satisfied for work to be done:

(i) a force should act on an object

(ii) the object must be displaced.

If any one of the above conditions does not exist, work is not done. This is the way we view work in science.

WORK DONE

➤ Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the applied force. 

Or

➤ We define work to be equal to the product of force and displacement.

Work done = force × displacement

`W = F s`

Or 

`W = Fs cos \theta`

■ Work has only magnitude and no direction.

If `F = 1 N` and `s = 1 m` then the work done by the force will be `1 N m`. Here the unit of work is newton metre `(N m)` or joule `(J)`.

ONE JOULE

➤ `1J` is the amount of work done on an object when a force of `1 N` displaces it by `1 m` along the line of action of the force.

Example (Work done)

A force of `5 N` is acting on an object. The object is displaced through `2 m` in the direction of the force (Fig). If the force acts on the object all through the displacement, then work done is `5 N × 2 m =10 N m` or `10 J`.

■ Work done is negative when the force acts opposite to the direction of displacement. Work done is positive when the force is in the direction of displacement.

Examples

(i) Work done is positive

A baby pulling a toy car parallel to the ground, as shown in Fig. The baby has exerted a force in the direction of displacement of the car. In this situation, the work done will be equal to the product of the force and displacement. In such situations, the work done by the force is taken as positive.

(ii)  Work done is negative 

Consider a situation in which an object is moving with a uniform velocity along a particular direction. Now a retarding force, `F`, is applied in the opposite direction. That is, the angle between the two directions is 180º. Let the object stop after a displacement `s`. In such a situation, the work done by the force, F is taken as negative and denoted by the minus sign. The work done by the force is `F × (–s) or (–F × s).`

Example (Numerical)

A porter lifts luggage of `15 kg` from the ground and puts it on his head `1.5 m` above the ground. Calculate the work done by him on the luggage.

Solution:

Mass of luggage, `m = 15 kg` and displacement, `s = 1.5 m.`

Work done, `W = F × s = mg × s`

`= 15 kg × 10  m s^{-2} × 1.5 m`

`= 225 kg  m s^{-2} m`

`= 225 N m = 225 J`

Work done is `225 J.`

ENERGY 

➤ An object having the capability to do work is said to possess energy. 

The object which does the work loses energy and the object on which the work is done gains energy. When this happens, energy is transferred from the former to the latter. The second object may move as it receives energy and therefore does some work. Thus, the first object had the capacity to do work. This implies that any object that possesses energy can do work.

The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule `(J)`. `1 J` is the energy required to do 1 joule of work.

FORMS OF ENERGY 

The various forms include mechanical energy (potential energy + kinetic energy), heat energy, chemical energy, electrical energy, and light energy.

KINETIC ENERGY 

Kinetic energy is the energy possessed by an object due to its motion. The kinetic energy of an object increases with its speed.

Express the kinetic energy of an object in the form of an equation

Consider an object of mass, `m` moving with a uniform velocity, `u`. Let it now be displaced through a distance `s` when a constant force, `F` acts on it in the direction of its displacement.  The work done, `W` is `F s`. The work done on the object will cause a change in its velocity. Let its velocity change from `u` to `v`. Let `a` be the acceleration produced. 

`v ^2 – u ^2 = 2a s`

`⇒ s = \frac{v^2-u^2}{2as}`  

Work done by the force `F`

`W = Fs`

`⇒ ma ( \frac{v^2-u^2}{2as})`

`⇒ \frac{1}{2}m (v^2-u^2)` 

If the object is starting from its stationary position, that is, `u = 0`, then

`⇒W= \frac{1}{2}m v^2`

👉It is clear that the work done is equal to the change in the kinetic energy of an object. 

`W = \Delta E_k`

👉Thus, the kinetic energy possessed by an object of mass, `m` and moving with a uniform velocity, `v` is

`E_k = \frac{1}{2}mv^2`