**CBSE NOTES CLASS 11 PHYSICS CHAPTER 6**

**WORK, ENERGY & POWER**

**Work**

Work is done by a force on the body over a certain displacement.

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

If we take the vectors $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{s}}$

W = $\overrightarrow{\mathrm{F}}$ . $\overrightarrow{\mathrm{s}}$ = F s cos θ, where θ is the angle between $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{s}}$.

No work is done if,

- the displacement is zero
- the force is zero.
- the force and displacement are mutually perpendicular.

For the block moving on a smooth horizontal table, the gravitational force *m g *does no work since it acts at right angles to the displacement.

Assuming the moon’s orbits around the earth is perfectly circular then the earth’s gravitational force does not do any work.

Work can be both positive and negative. If θ is between 0^{o} and 90^{o}, cos θ is positive. If θ is between 90^{o }and 180^{o}, cos θ is negative.

The frictional force opposes displacement and θ = 180^{o}, hence the work done by friction is negative (cos 180^{o} = –1).

**Kinetic Energy**

Energy of a body by virtue of its motion is called kinetic energy.

K = $\frac{1}{2}$ mv^{2}

**The Work Energy (WE) Theorem**

The change in kinetic energy of a particle is equal to the work done on it by the net force.

Work and energy have the **same dimensions**, [ML^{2}T^{–2}].

The SI unit of both work and energy is Joule** (J)**.

**Proof**

As per third equation of rectilinear motion,

v^{2}* − u*^{2}* = 2 a s*

where u and v are the initial and final speeds and s the distance traversed.

Multiplying both sides by m/2, we have

$\frac{1}{2}$* mv*^{2}* – *$\frac{1}{2}$* m u*^{2}* = m a s*

⇒* *$\frac{1}{2}$* mv*^{2}* – *$\frac{1}{2}$* mu*^{2}* = F s*

Where the last step is from Newton’s Second Law, i.e., F = *m* *a*.

Hence,

*K*_{f}* − K*_{i}* = W *

where K_{i}* *and K_{f}* *are the initial and final kinetic energies of the object, respectively.

**Work Done By a Variable Force**

Consider a plot of x and F(x).

We can divide the area into area elements of equal widths, Δ*x*.

If the displacement Δ*x *is small, we can take the force *F*(*x*) as approximately constant and the work done is then

Δ*W *= *F*(*x*) Δ*x*

Adding successive rectangular areas, we get the total work done as

$$\mathrm{W}\mathrm{}\cong \mathrm{}\underset{\mathit{\Delta x}\mathrm{}\mathrm{}0}{\mathit{lim}}{\mathrm{\Sigma}}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{}\mathrm{F}\left(\mathrm{x}\right)\mathrm{}\mathrm{\Delta}\mathrm{x}$$

If we take limit as displacements approach zero, then the sum approaches a definite value equal to the area under the curve or the integral,

$$\mathrm{}\mathrm{W}\mathrm{}=\mathrm{}{\int}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}\left(\mathrm{x}\right)\mathrm{}\mathrm{d}\mathrm{x}$$

**Work Energy Theorem for a Variable Force**

The time rate of change of kinetic energy is

$$\frac{\mathrm{d}\mathrm{K}}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}\left(\frac{1}{2}\mathrm{m}{\mathrm{v}}^{2}\right)=\mathrm{m}\mathrm{}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{t}}\mathrm{}\mathrm{v}=\mathrm{F}\mathrm{}\mathrm{v}$$

Thus

$$\frac{\mathrm{d}\mathrm{K}}{\mathrm{d}\mathrm{t}}=\mathrm{F}\left(\mathrm{x}\right)\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$$

*Or dK = F(x)* d*x*

Integrating from the initial position (*x*_{i}* *) to final position ( *x*_{f }), we have

$${\int}_{{\mathrm{K}}_{\mathrm{i}}}^{{\mathrm{K}}_{\mathrm{f}}}\mathrm{d}\mathrm{K}\mathrm{}={\int}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}\left(\mathrm{x}\right)\mathrm{}\mathrm{d}\mathrm{x}\mathrm{}$$

*⇒ K*_{f}* − K*_{i}* = W*

**Potential Energy**

Potential energy is the stored energy of an object. It is the energy by virtue of an object's position or configuration relative to other objects.

Potential energy is associated with restoring force.

Examples of potential energy are gravitational potential energy, electric potential energy energy stored in a stretched spring, etc.

Mathematically, the potential energy *V*(*x*) is defined if the force *F*(*x*) can be written as

$$\mathrm{F}\left(\mathrm{x}\right)\mathrm{}=\mathrm{}-\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{x}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}{\int}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}\left(\mathrm{x}\right)\mathrm{}\mathrm{d}\mathrm{x}=-{\int}_{{\mathrm{V}}_{\mathrm{i}}}^{{\mathrm{V}}_{\mathrm{f}}}\mathrm{}\mathrm{d}\mathrm{V}\mathrm{}=\mathrm{}{{\mathrm{V}}_{\mathrm{i}}-\mathrm{}\mathrm{V}}_{\mathrm{f}}\mathrm{}$$

The change in potential energy, for a conservative force, Δ*V *is equal to the negative of the work done by the force

Δ*V = *− *F*(*x*) Δ*x*

Gravitational potential energy

*V(h) = mgh, *

where h is the height of the object from the surface of the earth.

**Conservative force**

- A force
*F*(*x*) is conservative if it can be derived from a scalar quantity*V*(*x*) by the relation given by Δ*V =*−*F*(*x*) Δ*x.* - The work done by the conservative force depends only on the end points and not on the path followed.
*W = K*_{f }*– K*_{i }*= V*(*x*_{i}*V*(*x*_{f} - The work done by conservative force in a closed path is zero.
- We can find difference of Potential Energies between two points, but cannot find absolute Potential Energies.
- Time is not considered in the discussion.

**Conservation of Total Mechanical Energy**

The total mechanical energy of a system is conserved if the forces, doing work on it, are ** conservative**.

K + V = constant = Total mechanical energy

ΔK + ΔV = 0

Example – Consider a ball is dropped from height h.

TE at the top when the ball is at rest is PE = *mgh*

When the ball reaches the ground the PE becomes 0 and all the PE is converted to KE.

Just before hitting the ground,

$$\frac{1}{2}\mathrm{m}{\mathrm{v}}^{2}\mathrm{}=\mathrm{}\mathrm{m}\mathrm{g}\mathrm{h}\mathrm{}$$

*Or *v* = *$\mathrm{}\sqrt{2\mathrm{g}\mathrm{h}}$

**Example – Vertical circle**

The potential energy of the system is taken to be zero at the lowest point A.

At A, $$\mathrm{E}\mathrm{}=\frac{1}{2}\mathrm{m}\mathrm{}{{\mathrm{v}}_{\mathrm{A}}}^{2}$$ And $${\mathrm{T}}_{\mathrm{A}}\mathrm{}\u2013\mathrm{}\mathrm{m}\mathrm{g}\mathrm{}=\frac{\mathrm{m}{\mathrm{v}}^{2}}{\mathrm{r}}$$ At C, T |

And

$$\mathrm{E}\mathrm{}=\mathrm{}\frac{1}{2}\mathrm{m}\mathrm{}{{\mathrm{v}}_{\mathrm{c}}}^{2}\mathrm{}+\mathrm{}2\mathrm{}\mathrm{m}\mathrm{g}\mathrm{r}$$

Also** **

$$\mathrm{m}\mathrm{g}\mathrm{}=\frac{\mathrm{m}{{\mathrm{v}}_{\mathrm{c}}}^{2}}{\mathrm{r}}$$

$${\Rightarrow \mathrm{}\mathrm{v}}_{\mathrm{c}}\mathrm{}=\sqrt{\mathrm{g}\mathrm{r}}$$

$\Rightarrow \mathrm{}\mathrm{E}\mathrm{}=\mathrm{}\frac{5}{2}\mathrm{m}\mathrm{g}\mathrm{r}$* *

Comparing with the equation at **A**, we have

$$\frac{1}{2}\mathrm{m}\mathrm{}{{\mathrm{v}}_{\mathrm{A}}}^{2}=\frac{5}{2}\mathrm{}\mathrm{m}\mathrm{g}\mathrm{r}\mathrm{}$$

$${\Rightarrow \mathrm{}\mathrm{v}}_{\mathrm{A}}\mathrm{}=\sqrt{5\mathrm{g}\mathrm{r}}$$

At point B,

$$\mathrm{E}\mathrm{}=\mathrm{}\frac{1}{2}\mathrm{m}\mathrm{}{{\mathrm{v}}_{\mathrm{B}}}^{2}\mathrm{}+\mathrm{}\mathrm{m}\mathrm{g}\mathrm{r}$$

$$\frac{5}{2}\mathrm{}\mathrm{m}\mathrm{g}\mathrm{r}\mathrm{}\u2013\mathrm{m}\mathrm{g}\mathrm{r}\mathrm{}\mathrm{}=\frac{1}{2}\mathrm{m}\mathrm{}{{\mathrm{v}}_{\mathrm{B}}}^{2}\mathrm{}$$

$${\Rightarrow \mathrm{}\mathrm{v}}_{\mathrm{B}}\mathrm{}=\sqrt{3\mathrm{g}\mathrm{r}}$$

**The Potential Energy of a Stretched Spring**

The spring force is conservative. The system consists of a block attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall.

- The spring is considered to be massless.
- The surface has been considered to possess negligible friction.

Spring force is given by Hooke’s law

*F*_{s}* = *− *k x*

The constant *k *is called the spring constant. Its unit is N m^{-1}. The spring is said to be stiff if *k *is large and soft if *k *is small.

If the block is pulled outwards and the maximum extension is *x*_{m}, the work done by the spring (restoring) force is

$${\mathrm{W}}_{\mathrm{s}}=\mathrm{}{\int}_{0}^{{\mathrm{x}}_{\mathrm{m}}}\mathrm{F}\left(\mathrm{x}\right)\mathrm{}\mathrm{d}\mathrm{x}$$

$$=-\mathrm{}\mathrm{k}{\int}_{0}^{{\mathrm{x}}_{\mathrm{m}}}\mathrm{}\mathrm{x}\mathrm{}\mathrm{d}\mathrm{x}$$

$$\Rightarrow \mathrm{}{\mathrm{W}}_{\mathrm{s}}=\mathrm{}\frac{-\mathrm{k}{\mathrm{x}}_{\mathrm{m}}^{2}}{2}$$

The work done by the external pulling force *F *is positive.

The same is true when the spring is compressed with a displacement *x*_{c}* *(< 0).

If the block is moved from an initial displacement *x*_{i}* *to a final displacement *x*_{f}* *, the work done by the spring force *W _{s} *is

$${\mathrm{W}}_{\mathrm{s}}=\mathrm{}{\int}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}\left(\mathrm{x}\right)\mathrm{}\mathrm{d}\mathrm{x}$$

$$=-\mathrm{}\mathrm{k}{\int}_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{}\mathrm{x}\mathrm{}\mathrm{d}\mathrm{x}$$

$${\mathrm{W}}_{\mathrm{s}}=\frac{\mathrm{k}{\mathrm{x}}_{\mathrm{i}}^{2}}{2}-\frac{\mathrm{k}{\mathrm{x}}_{\mathrm{f}}^{2}}{2}$$

Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from x_{i} and allowed to return to x_{i}; W_{s} = 0.

⇒ Hence, the spring force is a **conservative force***.*

- The potential energy V(x) of the spring to be zero when block and spring system is in the equilibrium position. For an extension (or compression) x,
$$\mathrm{V}\mathrm{(x)}=\mathrm{}\frac{\mathrm{k}{\mathrm{x}}^{2}}{2}$$

- Total energy of the block,
$$\mathit{}\frac{\mathrm{k}{\mathrm{x}}_{\mathrm{m}}^{2}}{2}=\mathrm{}\frac{\mathrm{k}{\mathrm{x}}^{2}}{2}+\frac{1}{2}\mathrm{m}{\mathrm{v}}^{2}$$

When

*x = 0,*v = v_{m }$$\frac{\mathrm{k}{\mathrm{x}}_{\mathrm{m}}^{2}}{2}=\mathrm{}\frac{1}{2}\mathrm{m}{\mathrm{v}}_{\mathrm{m}}^{2}\mathrm{}\mathrm{}$$

$$\Rightarrow {\mathrm{v}}_{\mathrm{m}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{}\mathrm{}{\mathrm{x}}_{\mathrm{m}}$$

**Plots of the Potential Energy and Kinetic Energy of a Spring**

Plots of the potential energy V and kinetic energy K of a block attached to a spring obeying Hooke’s law, are parabolic. The two plots are complementary, one decreasing as the other increases. The total mechanical energy E = K + V remains constant.

**Work Done by Combination of Conservtive and Non-conservative Forces**

If the two forces on the body consist of a conservative force *F*_{c}* *and a non-conservative force *F*_{nc}, the conservation of mechanical energy formula will be,

(*F*_{c}*+ F*_{nc}* *) Δ*x = *Δ*K*

But *F*_{c}* *Δ*x = *− Δ*V*

Hence, Δ(*K + V*) *= F*_{nc}* *Δ*x*

Δ*E = F*_{nc}* *Δ*x*

*E*_{f}* *− *E*_{i}* = W*_{nc}

Where, *W*_{nc}* *is the total work done by the non-conservative forces over the path.

- Take up example 9

**Various forms of energy **

Energy exists in various forms which can transform into one another.

- Mechanical Energy – KE and PE
- Heat Energy
- Electrical Energy
- Light Energy
- Chemical Energy
- Nuclear Energy

**Conservation of Total Energy**

The total energy of the universe is constant. If one part of the universe loses energy, another part must gain an equal amount of energy.

The principle of conservation of energy cannot be proved. However, no violation of this principle has been observed.

**Power**

**Power **is defined as the time rate at which work is done or energy is transferred.

The average power of a force is defined as the ratio of the work, W, to the total time t taken

$${\mathrm{P}}_{\mathrm{a}\mathrm{v}}=\frac{\mathrm{W}}{\mathrm{t}}\mathrm{}$$

And

$${\mathrm{P}}_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{t}}=\frac{\mathrm{d}\mathrm{W}}{\mathrm{d}\mathrm{t}}$$

Work dW done by a force F for a displacement dr is, $\mathrm{d}\mathrm{W}\mathrm{}=\mathrm{}\overrightarrow{\mathrm{F}}.\mathrm{d}\overrightarrow{\mathrm{r}}$. The instantaneous power can therefore be expressed as

$${\mathrm{P}}_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{t}}=\overrightarrow{\mathrm{F}}.\frac{\mathrm{d}\overrightarrow{\mathrm{r}}}{\mathrm{d}\mathrm{t}}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{v}}$$

Power is a scalar quantity. Its dimensions are [ML^{2}T^{–3}].

The SI unit is called a watt (W).

- 1 watt = 1 Js
^{–1}. - 1 Horse-power (hp) = 746 W
- 1 kilowatt hour (kWh) of energy
= 1000 (watt).1 (hour) = 1000 watt hour

=1 kilowatt hour (kWh)

= 10

^{3}(W) × 3600 (s)= 3.6 × 10

^{6}J - Electrical energy is measured in units of kWh.

**Collisions**

Consider two masses m_{1} and m_{2}. The particle m_{1} is moving with speed u_{1} and m_{2} is at rest.

The mass m_{1 }collides with the stationary mass m_{2} and m_{1} and m_{2} move with velocities v_{1} and v_{2} respectively, after the collision. v_{1} and v_{2 }make angle θ_{1} and θ_{2} respectively with the original direction of u_{1}**.**

In all collisions the total linear momentum is conserved, that is, the initial momentum of the system is equal to the final momentum of the system.

Δp_{1} = F_{12} Δt

Δp_{2 }= F_{21} Δt

where F_{12} is the force exerted on the first particle by the second particle and F_{21} is the force exerted on the second particle by the first particle.

Now from Newton’s Third Law, F_{12} = − F_{21}. This implies

Δp_{1} *+ *Δp_{2} *= *0

The total kinetic energy of the system is not necessarily conserved. The impact and deformation during collision may generate heat and sound. Part of the initial kinetic energy is transformed into other forms of energy.

**Collisions in one dimension**

θ_{1} = θ_{2} = 0

**Inelastic collisions**

If the kinetic energy is not conserved, the collision is called inelastic collision. That is the KE before and after collision is not the same.

**Completely Inelastic Collision**

A collision in which the two particles move together after the collision is called a completely inelastic collision.

By conservation of momentum,

m_{1}u_{1 }= (m_{1 }+ m_{2})v [v is same for both]

$$\Rightarrow \mathrm{}\mathrm{v}=\mathrm{}\frac{{\mathrm{m}}_{1}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}\mathrm{}{\mathrm{u}}_{1}$$

The loss in kinetic energy on collision is

$$\mathrm{\Delta}\mathrm{K}\mathrm{}=\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{u}}_{1}^{2}-\frac{1}{2}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right){\mathrm{v}}^{2}$$

Putting the value of v,

$$=\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{u}}_{1}^{2}-\frac{1}{2}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{}\frac{{\mathrm{m}}_{1}^{2}{\mathrm{u}}_{1}^{2}}{{({\mathrm{m}}_{1}+{\mathrm{m}}_{2})}^{2}}$$

$$=\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{u}}_{1}^{2}(1-\mathrm{}\frac{{\mathrm{m}}_{1}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}})$$

$$=\frac{1}{2}\mathrm{}\left(\frac{{\mathrm{m}}_{1}{\mathrm{m}}_{2}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}\right){\mathrm{u}}_{1}^{2}$$

**Elastic Collisions**

If the kinetic energy is conserved, the collision is called elastic collision. That is the KE before and after collision is same.

By conservation of momentum,

m_{1}u_{1 }= m_{1}v_{1 }+ m_{2}v_{2}

⇒ m_{1 }(u_{1 }- v_{1})_{ }= m_{2}v_{2} -(1)

By conservation of kinetic energy,

$${\mathrm{m}}_{1}{\mathrm{u}}_{1}^{2}={\mathrm{m}}_{1}{\mathrm{v}}_{1}^{2}+\mathrm{}{\mathrm{m}}_{2}{\mathrm{v}}_{2}^{2}$$

$$\Rightarrow \mathrm{}\mathrm{}\mathrm{}{\mathrm{m}}_{1}\left({\mathrm{u}}_{1}^{2}-{\mathrm{v}}_{1}^{2}\right)=\mathrm{}{\mathrm{m}}_{2}{\mathrm{v}}_{2}^{2}\mathrm{}\mathrm{}\mathrm{}\mathrm{}-\left(2\right)$$

Dividing relation (2) by (1)

u_{1 }+ v_{1 }= v_{2}

Putting the value of v_{2}** **in (1)

$${\mathrm{v}}_{1}=\frac{{\mathrm{m}}_{1}-{\mathrm{m}}_{2}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}{\mathrm{u}}_{1}\mathrm{}$$

And

$${\mathrm{v}}_{2}=\frac{2{\mathrm{m}}_{1}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}}{\mathrm{u}}_{1}$$

**Case I: **

If the two masses are equal v_{1}= 0

v_{2} = u_{1}

The first mass comes to rest and pushes off the second mass with its initial speed on collision.

**Case II: **If one mass dominates, e.g. m_{2} > > m_{1}

v_{1} ~ − u_{1 }and_{ }v_{2} ~ 0

The heavier mass is undisturbed while the lighter mass reverses its velocity.

**Coefficient of restitution e**

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

The coefficient is related to energy by

$$\mathrm{e}=\sqrt{\frac{\mathrm{KE\; after\; the\; collision}}{\mathrm{KE\; before\; the\; collision}}}$$

**Range of values for e **

*e* is usually a positive, real number between 0 and 1.

*e* = 0: This is a perfectly inelastic collision. The objects do not move apart after the collision, but instead they coalesce. Kinetic energy is converted to heat or work done in deforming the objects.

0 < *e* < 1: This is a real-world inelastic collision, in which some kinetic energy is dissipated.

*e* = 1: This is a perfectly elastic collision, in which no kinetic energy is dissipated, and the objects rebound from one another with the same relative speed with which they approached.

*e* < 0: A collision in which the separation velocity of the objects has the same direction (sign) as the closing velocity, implying the objects passed through one another without fully engaging. This may also be thought of as an incomplete transfer of momentum. An example of this might be a small, dense object passing through a large, less dense one – e.g., a bullet passing through a target, or a motorcycle passing through a motor home or a wave tearing through a dam.

**Collisions in Two Dimensions**

Conserving the momentum in x - direction,

m_{1}u_{1} = m_{1}v_{1} cos θ_{1} + m_{2}v_{2} cos θ_{2}

Conserving the momentum in y - direction,

0 = m_{1}v_{1} sin θ_{1} − m_{2}v_{2} sin θ_{2}

If the collision is elastic then,

$${\mathrm{m}}_{1}{\mathrm{u}}_{1}^{2}={\mathrm{m}}_{1}{\mathrm{v}}_{1}^{2}+\mathrm{}{\mathrm{m}}_{2}{\mathrm{v}}_{2}^{2}$$

Now we have four unknowns (v_{1}, v_{2}, θ_{1}, θ_{2}) but only 3 equations. If one of the variables is known, we can solve for other 3 variables.