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# Physics - 11 Complete Syllabus Notes

### Chapter 4 - Work Energy and Power

Work :
Work is said to be done by a force on a body, if the body moves in any direction except at 900 from force.

It is defines as the scalar product of force vector and displacement vector.
i.e. $W&space;=&space;\overrightarrow{F}\times&space;\overrightarrow{S}$
W = FScos $\theta$
Where $\theta$ is anbgle between $\overrightarrow{F}&space;and&space;\overrightarrow{S}$

It is scalar quantity. It's SI unit is Joule (Kgm 2s-2) and cgs unit is (gcm 2s-2). It's dimensional formula is [ML 2T-2]

i.e. w = Fscos $\theta$
= positive works

For eg: When oject is falling freely under the action of gravity than work done by gravitational force an the object is an example of positive work.

If $\theta$ = 900 then work done wil be zero

i.e. w = Fscos900
w = zero work

For eg: When a man carrying a load on his head and moving on a horizontalroad then work done by man on the load is an example of zero workdone.

i.e. w = Fscos1800
w = Negative work

For eg: When object is thrown verticaly upward against the gravity then workdone by gravitational force on the object is an example of negative work.

Energy:
The capacity of doing work is called energy. It is scalar quantity and its unit in SI syste is joule and cgs is erg. Dimensional formula is [ML 2T-2]
1 J = 107 erg.

Various Forms Of Energy are:

• Heat Energy
• Light Energy
• Electrical Energy
• Mechanical Energy(Kinetic Energy + Potential Energy)

• Kinetic Energy:

The energy possess by an object due to it's motion is called Kinetic Energy.
Expression For K.E.

Consider an object mass "m " is initially at rest. Suppose a constant force 'F' is applied on object to produce a displacement $\overrightarrow{F}$
Now, work done by constant force,
W = $\overrightarrow{F}$ . $\overrightarrow{S}$
W = Fscosϴ (Where ϴ is angle between F and S vector)
W = Fscos0
W = Fs
W = mas ..........i)

F = ma where a is the acceleration of object. If v be the final speed attained by an object then from equation of motion v2 = u2 + 2as we ca write:
v 2 = 02 + 2as (since u=0)
v 2 = 2as
v 2 / 2 = as ..........ii)

From eqn i) and ii), we get,
w = m × v2 / s
w = 1/2 mv2

This work done is equal to kinetic energy
Kinetic energy = 1/2 mv2 .........iii)

Work Energy Theorem

The total work done by a force acting on a body is equal to total change in it's kinetic energy.
i.e, Total workdone = Total change in knetic energy

Proof:,

Consider an object mass 'm' is moving with constant speed 'u'.
Suppose a constant force F vector is applied on object is produce a displacement S vector.
Now, workdone by constant force
W = $\overrightarrow{F}$ . $\overrightarrow{S}$
W = Fscosϴ (Where ϴ is angle between F and S vector)
W = Fscos0
W = Fs
W = mas .......i) (since f = ma where 'a' is acceleration of a object)

If v be the final speed attained by an object then from equation of motion v 2 = u2 + 2as
$\frac{v^{2}-u^{2}}{2}=&space;as\rightarrow&space;ii)$
Now from eqn i) and ii)

$W&space;=m\times&space;\left&space;(&space;\frac{v^{2}-u^{2}}{2}&space;\right&space;)$
W = 1/2 mv2 - 1/2 mu2

Total work done = final K.E - Initial K.E
Therefore, Total work done = Total change in kinetic energy which is work energy theorem.

Potential energy:
The energy possesses by an object due to its position or configuration is called potential energy.

Gravitational Potential Energy = mgh
Elastic Potential Energy = 1/2 kx2

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