# Work, Energy and Power Class 11 Physics | Notes

## Work

Work is said to be done if a body gets displaced in the direction of applied force. It is a scalar quantity. Its unit is Joule (J) in S.I. and erg in C.G.S. It is also defined as the dot product of force ($\overrightarrow{F}$) and displacement ($\overrightarrow{S}$),

i.e. W =  $\overrightarrow{F}$.$\overrightarrow{S}$

$\therefore$  W = FS Cos$\theta$ where $\theta$ is the angle between $\overrightarrow{F}$and $\overrightarrow{S}$.

Special cases:

1. When $\theta$ = 0º,

w = FScos$\theta$ = FScos0º = FS (+ve)

The positive sign shows that the work is done in the direction of applied force.

2. When $\theta$ = 180º

w = FScos180º = –FS (–ve)

The negative sign shows that the work is done in the direction opposite to applied force.

3. When $\theta$ = 90º

w = FScos90º = 0 (+ve)

The +ve sign shows that the work is done in the direction of applied force.

## Energy

The capacity of doing work is known as energy. It’s unit is Joule (J) in S.I. and erg in C.G.S.

1J = 107 erg

### Types of energy

#### 1.Kinetic Energy (K.E.)

It is the energy possessed by a body due to its motion. It is given by K.E. = $\frac{1}{2}$ mv2.

#### 2.Potential Energy (P.E.)

It is the energy possessed by a body due to its position. It is given by P.E. = mgh

### Mechanical energy

It is the sum of kinetic energy (K.E.) and potential energy (P.E.)

## Power:

The rate of doing work is called power. It is given by

P = $\frac{w}{t}$ = $\frac{\overrightarrow{F}.\overrightarrow{S}}{t}$ = $\overrightarrow{F}$. $\frac{\overrightarrow{S}}{t}$ =$\overrightarrow{F}$ . $\overrightarrow{V}$ = FVcos$\theta$

If the displacement is produced is the direction of applied force, i.e. $\theta$ = 0º,

then P = $\frac{w}{t}$ = FV

Its unit is Js–1 or watt (w)

1 K.W = 103w

1MW = 106w

1 H.P. = 746 w

1mW = 10–3w

1$\mu$W = 10–6w

1nw = 10–9w

## Relation between K.E. and linear momentum (P)

Let us consider a body of mass ‘m’ moving with velocity ‘v’. Then we have

Linear momentum (P) = mv

and K.E. = $\frac{1}{2}$ mv2

= $\frac{1}{2}$ mv2$\times $$\frac{m}{m} = \frac{1}{2} \frac{{{m}^{2}}{{v}^{2}}}{m} =\frac{1}{2}$$\frac{{{(mv)}^{2}}}{m}$

$\therefore$ K.E. = $\frac{{{P}^{2}}}{2m}$

## Work – energy Theorem:

Work-energy theorem states that the change in kinetic energy of a body is equal to work done. Let us consider a body of mass ‘m’ moving with initial velocity ‘u’ at a point. If force ‘F’ is applied on it, it moves with constant acceleration ‘a’. After time ‘t’, its velocity becomes ‘v’ at point B. Let ‘S’ be the displacement covered by the body in time ‘t’.

Then, the amount of work done on moving from A to B is

W = Fs

Or, W = mas …………… (i)   [F = ma]

We have,

v2 = u2 + 2as

Or, s = $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$ …… (ii)

From (i) and (ii),

W = ma $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$

Or, W = $\frac{1}{2}$ mv2 –$\frac{1}{2}$ mu2

Or, W = (K.E)final – (K.E)initial

$\therefore$W = $\Delta$K.E

which is the work energy theorem.

If the body is moving in vertical direction under the action of gravity, then

$\Delta$K.E. = $\Delta$P.E.

$\therefore$W = $\Delta$K.E. = $\Delta$P.E.

Formulae

1. W = FScos$\theta$

2. P = $\frac{W}{t}$ = FVcos$\theta$

For $\theta$= 0O , P = Force× Velocity

3. K.E = $\frac{{{P}^{2}}}{2m}$

4. W = $\Delta$K.E = $\Delta$P.E

## Work done by Variable force

A force is said to be variable if it changes with respect to position. Let us consider a variable force F(x) is applied on a body and the body displaces from A to B in a fixed direction (X-axis). We can consider the entire displacement (AB) as a sum of a number of infinitesimal displacements. Let PQ = dx is one of the infinitesimal displacements. Then the small amount of work done in moving the body from P to Q is given by

dw = F × dx = PS × PQ

or, dw = Area of rectangle PQTS

For dx$\to$ 0, ST $\approx$ $\overset\frown{SR}$

∴ dw = area of strip PQRS ………… (i)

Now, the total work done in moving the object from A to B can be obtained by integrating equation (i) from x = x1 to x = x2, we get,

w = $\int\limits_{x1}^{x2}{dw}$ = $\int\limits_{x1}^{x2}{area\text{ }of\text{ }strip\text{ }PQRS}$

$\therefore$ w = area of ABCDA

This shows that the work done by the variable force is numerically equal to the area between force curve and displacement axis.

## Work done by constant force

Force is said to be constant if it does not change with respect to position. Let us consider a body of mass ‘m’ moving with initial velocity ‘u’ at a point. If force ‘F’ is applied on it, it moves with constant acceleration ‘a’. After time ‘t’, its velocity becomes ‘v’ at point B. Let ‘S’ be the displacement covered by the body in time ‘t’.

Then, the amount of work done on moving from A to B is

W = Fs

Or, W = ma s …………… (i)   [$\because$F = ma]

We have,

v2 = u2 + 2as

Or, s = $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$ ………… (ii)

From (i) and (ii),

W = ma $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$

Or, W = $\frac{1}{2}$ mv2 –$\frac{1}{2}$ mu2

Or, W = (K.E)final – (K.E)initial

$\therefore$W = $\Delta$K.E

Hence, Work done by constant force is equal to change in kinetic energy of the body.

## Principle of Conservation of energy (Mechanical):

Principle of Conservation of energy states that energy can neither be created nor destroyed, but can be changed from one form to another form.

For a freely falling body, the total mechanical energy of the body remains constant throughout the motion. Let us consider an object of mass ‘m’ is initially at rest at point A, which is at a height of ‘h’ from the ground level. The object is allowed to fall freely under the action of gravity. After covering a distance ‘x’, it reaches point B and finally reaches the ground level at point C.

At position A

Height = h and velocity = VA = 0

$\therefore$ K.EA = $\frac{1}{2}$ mVA2 = 0

And P.EA = mgh

Now, mechanical energy at point A = (M.E.)A = K.EA + P.EA

= 0 + mgh

= mgh

$\therefore$ M.EA = mgh ……….. (i)

At position B

Height = h – x and velocity = VB

We know, VB2 = VA2 + 2aS

or, VB2 = VA2 + 2ghx

= 0 + 2gx

Now, K.EB = $\frac{1}{2}$m (VB)2

= $\frac{1}{2}$m × 2gx = mgx

P.EB = mg(h – x)

Now, mechanical energy at B, (M.E)B = K.EB + P.EB

= mgx + mg(h – x)

= mgx + mgh – mgx

= mgh

$\therefore$ M.EB = mgh ………… (ii)

At position C

Height = 0 and velocity = VC

We know,

VC2 = VA2 + 2gS

or, VC2 = 0 + 2gh

Now, K.EC = $\frac{1}{2}$ m(VC)2 = $\frac{1}{2}$m × 2gh = mgh

P.EC = mgh × 0 = 0

Now, mechanical energy at C, M.EC = K.EC + P.EC

= mgh

$\therefore$ M.EC = mgh ……… (iii)

From (i), (ii) and (iii)

M.EA = M.EB = M.EC

This shows that for a freely falling body, the total energy (mechanical) remains constant throughout the motion. Fig: Variation of K.E. and P.E. with height

## One dimensional collision

### a.Elastic collision

A collision is said to be elastic if both linear momentum (P) and kinetic energy (K.E.) remain conserved. There is no loss of energy in elastic collisions. Let us consider two objects of masses ‘m1‘ and ‘m2‘ moving with velocities ‘u1‘ and ‘u2‘ respectively. Let ‘u1‘ is greater than ‘u2‘, then the two objects collide and more separately with final velocities ‘v1‘ and ‘v2‘ respectively. Since the collision in elastic, from conservation of linear momentum.

m1u1 + m2u2 = m1v1 + m2v2

or, m1(u1 – v1) = m2(v2 – u2) ……….. (i)

And from conservation of kinetic energy,

$\frac{1}{2}$m1u12 +$\frac{1}{2}$ m2u22  =$\frac{1}{2}$ m1v12 +$\frac{1}{2}$ m2v22

or, m1(u12 – v12) = m2(u22 – v22) ………. (ii)

Dividing (ii) by (i),

u1 + v1 = u2 + v2

or, u1 – u2 = v2 – v1

$\therefore$ u1 – u2 = v2 – v1

Above result shows that velocity of approach is equal to the velocity of separation in elastic collision, i.e. relative velocity of before collision is equal to that after collision.

Further,

v1 = v2 + u2 – u1 ………. (iii)

And, v2 = v1 + u1 – u2 ……….. (iv)

From equation (i) and (iii),

m1[u1 – (v2 + u2 – u1)] = m2(v2 – u2)

or, m1[u1 – v2 – u2 + u1) = m2v2 – m2u2

or, 2m1u1 – m1u2 – m1v2 = m2v2 – m2u

or, 2m1u1 – m1u2 + m2u2 = m1v2 – m2v

or, 2m1u1 – u2(m1 – m2) = v2(m1 + m2)

or, v2 = $\frac{2{{m}_{1}}u{}_{1}+{{u}_{2}}({{m}_{2}}-{{m}_{1}})}{{{m}_{1}}+{{m}_{2}}}$……….. (v)

From (i) and (iv),

m1(u1 – v1) = m2[(v1 + u1 – u2) – u2]

or, m1u1 – m1v1 = m2v1 + m2u1 – 2m2u2

or, m21 – m2u1 + 2m2u2 = m2v1 + m1v1

or, 2m2u2 + u1 (m1 – m2) = v1(m1 + m2)

or, v1 =$\frac{2{{m}_{2}}u{}_{2}+{{u}_{1}}({{m}_{1}}-{{m}_{2}})}{{{m}_{1}}+{{m}_{2}}}$ ……….. (vi)

From (v) and (vi) give the final velocities of m2 and m1 respectively.

Special cases

1. Case I

When m1 = m2 = m (Say).

Then from equations (v) and (vi), we get,

v1 = $\frac{2m{{u}_{2}}}{2m}$

$\therefore$ v1 = u2……….. (vii)

And, v2 = $\frac{2m{{u}_{1}}}{2m}$

$\therefore$ v2 = u1……….. (viii)

Equations (vii) and (viii) show that in elastic collisions the colliding bodies exchange their velocities if they have the same masses.

2. Case II

When m1 >>>> m2 and u2 = 0

Then from equations (v) and (vi), we get,

v1 $\approx$ $\frac{0+{{u}_{1}}{{m}_{1}}}{{{m}_{1}}}$

$\therefore$v1$\approx$ u1

And, v2 $\approx$ 2u1

3. Case III

When m2 >>>> m1 and u2 = 0

Then from equations (v) and (vi), we get,

v1 $\approx$ – u1

And v2 $\approx$ 0

Q.1. Show that in one dimensional elastic collision, velocity of approach is equal to velocity of separation.

i.e. [u1 – u2 = v2 – v1]

Q.2. Show that in one dimensional elastic collision, the colliding bodies exchange their velocities (if they have the same masses). i.e. [u1 = v2 and u2 = v1]

### b Inelastic collision

The collision is said to be inelastic if linear momentum is conserved but not the kinetic energy. In inelastic collision, there is loss of kinetic energy. Let us consider an object of mass ‘m1‘ moving with initial velocity ‘u1‘ collides with another object of mass ‘m2‘, which is initially at rest (i.e. u2 = 0). After collision, the colliding bodies combine and move together with common velocity ‘v’. Since the collision is inelastic, only linear momentum is conserved, not the kinetic energy.

From the conservation of linear momentum,

m1u1 + m2u2 = m1v + m2v

Or, m1u1 + 0 = (m1 + m2)v [$\because$ m2 is initially at rest]

$\therefore$ v = $\frac{{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}$……… (i)

The sum of K.E. before collision is

K.E.1 = $\frac{1}{2}$m1u12 + $\frac{1}{2}$m1u22

$\therefore$ K.E.1 = $\frac{1}{2}$m1u12  …. (ii)

Now, the sum of K.E. after collision is

K.E.2 =  $\frac{1}{2}$m1v2 + $\frac{1}{2}$m2v2

$\therefore$ K.E.2 =  $\frac{1}{2}$(m1 + m2)v2…… (iii)

Dividing (ii) by (iii)

$\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{({{m}_{1}}+{{m}_{2}}){{v}^{2}}}{{{m}_{1}}{{u}_{1}}^{2}}$

Using equation (i),

$\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{({{m}_{1}}+{{m}_{2}})}{{{m}_{1}}{{u}_{1}}^{2}}{{(\frac{{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}})}^{2}}$

Or, $\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}$

Or, $\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ < 1

$\therefore$K.E.2 < K.E.1

Above relation shows that the kinetic energy after collision is less than that before energy, i.e. there is loss in kinetic energy in inelastic collision.

## Conservative Force

A force is said to be conservative if work done by it in a closed path (loop) is zero. The work done by conservative force is path independent.

e.g. gravitational force, electrostatic force, magnetic force.

## Non–conservative Force

A force is said to be non – conservative if work done by it in a closed path (loop) is not zero. The work done by non – conservative force is path dependent.

e.g. viscous force, frictional force, etc.

## Coefficient of Restitution (e):

It is defined as the ratio of velocity of separation to the velocity of approaches. It is denoted by ‘e’ and given by

e = $\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}$

(i) For perfectly elastic collision, e = 1

(ii) For perfectly inelastic collision, e = 0

(iii) Practice in collision, 0 < e < 1

Also Read: Laws of Motion Class 11