# Work, Energy and Power

### Work done

Work done by a force is the product of the forces and the distance moved in the direction of the force.
SI unit: joule (J) – the SI unit of energy.
1 J = 1 N m

\begin{aligned} W=F\times s \end{aligned}

Work done must be calculated by using the component of the force that is parallel to the direction of the displacement.
If the two directions are the same, work is done on the object and the objects gains energy.
If the two directions are opposite, then energy is lost through work done(typically as friction or air resistance)

### Work done by a gas

Work done by a gas equals to the pressure it exerts over the cross-sectional area and the displacement the piston moves.

\begin{aligned} F&=p\times A\\W&=p\times A\times s\\&=p\times \Delta V \end{aligned}

### Gravitational potential energy

Work is done by you during lifting. Object gains height when work is done against gravity. Gravitational potential energy increases as an object moves to a higher ground.

### Other forms of potential energy

There are many forms of potential energy. Some examples are below:

1. Electrical potential energy – capacitor. A capacitor stores charge which provides the capacitor the ability to produce current. A battery does not store electrical potential energy. Instead, it is stored as chemical energy.
2. Elastic potential energy – spring
3. Chemical potential energy – food, battery. Usually, chemical energy allows the substance to burn for a period of time, such as most organic compounds.
4. Nuclear energy – radioactive nuclei

### Kinetic energy

Kinetic energy is the energy due to motion. A moving object contains kinetic energy. To derive the equation of kinetic energy, we first assume an object is at rest. A constant force does work on the object, causing it to accelerate.

\begin{aligned} v^2&=u^2+2as\\&=2as \text{ since the object is initially at rest} \end{aligned}

Multiplying $\frac{1}{2}m$ on both sides,

\begin{aligned} \frac{1}{2}mv^2&=mas\\&=Fs \end{aligned}

Hence, work done is converted into kinetic energy of $\frac{1}{2}mv^2$ .

### Gravitational PE and Kinetic energy Transformation

When a high object falls to a lower height, the GPE is converted to KE. The increase in KE cause the object to increase in speed. The decrease in height should always be calculated based on the drop in vertical height.

### Efficiency

Efficiency is the ratio of the useful output energy to the total input energy expressed as a percentage.

\begin{aligned} \text{efficiency}=\frac{\text{useful output energy}}{\text{total input energy}} \times 100\% \end{aligned}

It is important for you to determine what makes the useful output energy in an energy transformation situation. Below are some situations of energy transformations. Suggest the useful output energy and the energy/energies that are lost.

1. A luggage being delivered from the ground floor to the first floor.
2. A man turning a generator to produce electricity.
3. Water flowing through a hydroelectric damp.
4. Wind flowing through a windmill.

# Coulomb’s Law

Coulomb’s force is the electrical force between two charges. Since there are positive and negative charges, coulomb’s force can be attractive or repulsive. However, the magnitude of the force produced by a point charge is similar to gravitational force:

\begin{aligned} F=\pm \frac{1}{4\pi \epsilon _0} \frac{Q_1Q_2}{r^2} \end{aligned}

$\epsilon_o$ is the permittivity of free space. It is a description of how well electric field lines permeates in vacuum. It has a value of $8.85\times 10^{-12} \text{m}^{-3}\text{ kg}^{-1}\text{s}^4\text{A}^2$

### Electric Field

An electric field is a region which a charge experiences an electrical force.

Similar to gravitational field strength, the electric field strength at a point is the coulomb force per unit charge placed at that point.

\begin{aligned} E=\pm \frac{1}{4\pi \epsilon _0}\frac{Q}{r^2} \end{aligned}

Properties of electric field

• Field lines are represented by arrows, starting from positive and ends at the negative.
• Field lines do not cross.
• Field lines always emerge or enters a conducting surface(s.g. metal) at a perpendicular direction.
• There are no electric field in a conductor.

The direction of the field line represents the direction of electric force acting on a positive charge when it is placed in the field. As such, the direction of electric force on a negative charge is opposite to the direction of the field lines.

### Challenge 1

Can you suggest why the electric field lines emerging or entering a conducting surface is perpendicular?

### Electric Potential

Electric potential of a point is the amount of work done in bringing a unit charge from infinity to that point.

Similar to gravitational potential,

\begin{aligned} \phi=\pm \frac{1}{4 \pi \epsilon_o}\frac{Q}{r} \end{aligned}

The electric potential energy is the amount of work done in bring a charge from infinity to that point.

The difference between electric potential and potential energy is that for electric potential, it is the energy per unit charge while in potential energy, we are considering the entire charge, which may be more than 1 coulomb.

### Relationship between electric field and potential

It is important to remember that the field strength is the negative of the potential gradient. Potential gradient means that gradient of the potential-distance graph.

### Uniform Electric Field

A uniform electric field can be produced by a pair of parallel plates. A charge placed anywhere inside a uniform field experiences the same force, regardless of whether it is nearer to the positive or negative plate.

You can think that a positive charge, if placed near the negative plate, experience more attraction by the negative plate and less repulsion by the positive plate. The same charge placed near the positive plate experiences more repulsion from the positive plate and less attraction by the negative plate. Hence the charge experiences the same force anywhere in this uniform field.

# Energy of a Simple Harmonic Oscillator

### Energy changes in according to displacement

The kinetic energy of a simple harmonic oscillator is

\begin{aligned} E_\text{kinetic} &= \frac{1}{2}m \omega^2(x_o^2 - x^2) \end{aligned}

In a SHM, the oscillator’s kinetic energy and potential energy always changes from maximum to zero throughout the oscillations. However, at all time, the total energy of the oscillator is constant. This value can be obtained by calculating the maximum kinetic energy of the system:

\begin{aligned} E_\text{total} &= \frac{1}{2}mv_o ^2\\&=\frac{1}{2}m \omega^2 x_o^2 \end{aligned}

To find out the potential energy of a simple harmonic oscillator,

\begin{aligned} E_\text{potential} &= E_\text{total} - E_\text{kinetic}\\&=\frac{1}{2}m \omega^2x_o^2 - \frac{1}{2}m \omega^2(x_o^2 - x^2)\\&=\frac{1}{2}m\omega^2 x^2\end{aligned}

The energy changes of an SHM oscillator changes in a sinusoidal pattern.

### Energy changes according to time

From the velocity equation, the kinetic energy is

\begin{aligned} E_\text{kinetic}&=\frac{1}{2}mv^2\\&=\frac{1}{2}mv_o ^2 \cos^2{\omega t}\end{aligned}

The potential energy is the difference between the total energy and the kinetic energy,

\begin{aligned} E_\text{potential}&=\frac{1}{2}mv_o ^2 - \frac{1}{2}mv_o ^2 \cos^2{\omega t}\\&=\frac{1}{2}mv_o^2(1-\cos^2{\omega t})\\&=\frac{1}{2}mv_o \sin^2{\omega t} \end{aligned}

# Uniform Gravitational Force

### Uniform gravitational force

The gravitational force acting on a mass is called weight. Weight always acts vertically down and its magnitude is directly proportional to the mass.

\begin{aligned} W=mg \end{aligned}

where g is the gravitational field strength on the Earth’s surface. Gravitational field strength of a point is the gravitational force acting on a unit mass at that point.

\begin{aligned} g=\frac{F}{m} \end{aligned}

The SI unit of gravitational field strength is $\text{N kg}^{-1}$ .

On the Earth’s surface, the gravitational field is uniform and its direction is vertically down. A uniform field is represented by parallel field lines. A mass placed anywhere in a uniform field experiences the same gravitational force.

The direction of the gravitational field indicates the direction of the force acting on a mass placed in the field.

When a mass is thrown vertically upward, its initial velocity is v. At its maximum height, its velocity is $0\text{ m s}^{-1}$. It then falls and its velocity reaches v in the downward direction.

#### Challenge 1

What are the values of acceleration at point A, B and C?

# The Physics of Simple Harmonic Motion

An object moving in an SHM must obey the relation

\begin{aligned} a=-\omega ^2 x\end{aligned}

where $x$ is the displacement and $\omega$ is the angular frequency of the oscillator. $\omega$  is related to the frequency $f$ of the oscillator by the relation \begin{aligned} w=2\pi f \end{aligned} .

A body is moving in SHM if its acceleration is directly proportional to the displacement and in the opposite direction to the displacement.

In this equation, we observe the following:

• the acceleration of the oscillator is always opposite to the displacement,
• the magnitude of the acceleration is directly proportional to the displacement.

From these two points, we see that the acceleration is greatest when the oscillator is at the maximum displacement while it is zero when the oscillator is at equilibrium. Since force and acceleration is directly proportional $F=ma$ , we see that the restoring force of the oscillator is correspondingly largest at the maximum displacement and zero at equilibrium.

### SHM in a spring

When a object hung on a spring oscillates, it is moving in SHM. There is a restoring force from the spring causing the object to move up and down. There are two ways to start the motion:

1. displace the object to a certain distance and release, or
2. with the oscillator at equilibrium, give it a push such that it starts moving.

Which colour of the graphs would case 1 and 2 belong?

The red graph would have an equation of \begin{aligned} x=x_0 \sin\omega t \end{aligned} while the blue graph is \begin{aligned} x=x_0 \cos \omega t \end{aligned}.

For the rest of this post, I will use the first equation (red), \begin{aligned} x = x_0 \sin \omega t \end{aligned}.

Since velocity is the derivative of displacement with time,

\begin{aligned} v=\frac{dx}{dt}=x_0 \omega \cos \omega t \end{aligned}

Acceleration is the derivative of velocity with time, hence

\begin{aligned} a=\frac{dv}{dt}= x_0 \omega^2 \sin \omega t\end{aligned}

But from the first equation $x = x_0 \sin \omega t$ , we have

\begin{aligned} a = -x_0 \omega^2 \end{aligned}

### Equation relating velocity to displacement

In the above equations, we have relationships with respect to time. We can also determine the velocity of the object if the displacement is known.

From the original displacement-time equation,

\begin{aligned} x&=x_0 \sin{\omega t} \end{aligned}

\begin{aligned} v=x_0 \omega \cos{\omega t} \end{aligned}

Squaring both equations, we have

\begin{aligned} x^2&=x_0 ^2 \sin^2{\omega t}\\v^2&=x_0 ^2 \omega ^2\cos^2{\omega t}\\&=x_0 ^2 \omega^2 (1-\sin^2{\omega t})\\&=x_0 ^2 \omega^2(1-\frac{x^2}{x_0 ^2})\end{aligned}

Simplifying,

\begin{aligned} v^2 &= \omega^2(x_0 ^2 - x^2) \end{aligned}

Hence,

\begin{aligned} v = \pm\omega \sqrt(x_0 ^2 - x^2) \end{aligned}

### The Restoring Force

It is important to understand that the restoring force is the resultant of both the spring force and the weight of the oscillator. Since weight is constant, the restoring force would be proportional to the spring force, $F=-kx$. Hence, the restoring force is never constant.

\begin{aligned} F_R=F_S-w \end{aligned}

where $F_R$ is the restoring force and $F_S$ is the spring force.

From the equation, it can be seen that the maximum restoring force occurs when $F_S$ is maximum, which occurs when the displacement is at maximum. This is also the position where the acceleration is maximum(and zero velocity).

### Phases of an SHM

\begin{aligned} x &= x_0 \sin \omega t \\ v &= x_0 \omega \cos \omega t\\ a &= -x_0 \omega ^2 \sin \omega t\end{aligned}

As we can see,

• the velocity of SHM $v$ and displacement $x$ has a phase difference of \begin{aligned} \frac{\pi}{2} \end{aligned} .
• The acceleration $a$ and the displacement has a phase difference of $\pi$.
• The acceleration and velocity has a phase difference of $\frac{\pi}{2}$

### Review Question 1

Using the blue graph above, sketch the corresponding graphs of $v$ and $a$.

# Simple Harmonic Motion and Circular Motion

Simple harmonic motion is closely related to a uniform circular motion.

A real life simple harmonic motion.

### Review Question 1

Answer the following questions as you watch the first YouTube video(Simple harmonic motion and uniform circular motion).

1. How does the centre of the red cylinder and the green ball relate?
2. At which part of the SHM(left) is the red cylinder moving the fastest and slowest?
3. At which part of the circular motion(right) is the green ball moving vertically the fastest and slowest?

### Review Question 2

Navigate to this site. This animation shows how a 4-stroke engine internal combustion works.

Answer the following questions:

1. The piston is moving with SHM. At which point is the piston the fastest?
2. At which point is the piston moving the slowest?

Given that the maximum velocity of the piston is $v$,

1. write down the expression of the velocity of the piston at the position shown in the diagram.
2. write down the expression of the acceleration of the piston at the position shown in the diagram.
3. Sketch a velocity-time graph of the SHM demonstrated by the piston. Draw two complete cycles and label the parts A, B and C on your graph.

# Mathematics of Circular Motion

### Angular Displacement

The angle which an object moves around a circle is called the angular displacement. This angle is usually expressed in radians.

\begin{aligned} \pi = 180^\circ \end{aligned}

Hence, there are $2 \pi$ radians in one complete circle.

The relationship between the angle in radian and the circle is

\begin{aligned} \theta =\frac{s}{r} \end{aligned}

where $s$ is the angular displacement in radians, and $r$ is the radius of the circle. The unit of $\omega$ is $\text{rad s}^{-1}$ .

### Angular Velocity

Angular velocity is the rate of change of angular displacement.

Mathematically, angular velocity $\omega$ is

\begin{aligned} \omega = \frac{\Delta \theta}{\Delta t} \end{aligned}

Since an object moves one circumference in one period time, we have

\begin{aligned} \omega &= \frac{2 \pi}{T}\\v &= \frac{2 \pi r}{T}\end{aligned}

Eliminating $T$ ,

\begin{aligned} v=r\omega \end{aligned}

### Review

Question 1

An object moves round a circle of 15 cm radius at a constant speed. It completes one revolution in 8.0 s. Calculate its angular velocity and its velocity.

Solution

\begin{aligned} \omega &= \frac{2 \pi}{8.0}\\&= 0.785\text{ rad s}^{-1}\end{aligned} \begin{aligned} v &= r \omega\\&= 15 \times 0.785\\&= 11.8\text{ cm s}^{-1}\end{aligned}

# How a Capacitor Discharge

### How does a capacitor work?

When a capacitor is connected across a power source, it starts charging up as current flows in the circuit.

Initially when the capacitor has not stored any charge, its potential difference is 0 V. As current flows, charges start storing across its plates, and the potential difference increases. The current also starts to decrease because the capacitor reduces the overall e.m.f of the circuit. When the potential difference of the capacitor equals to the e.m.f. of the power source, current stops flowing.

After the capacitor is fully charged, the switches are set to the figure above. The capacitor starts discharging and a current flows in the left part of the circuit. The potential difference across the capacitor follows an exponential curve.

### Wolfram Demonstration

Click this Wolfram Demonstration Project link to access the simulation. This simulation shows how the potential difference across the capacitor decreases during a discharge. The rate of discharge is dependent on the capacitance of the capacitor. A larger capacitor stores more charge. Although the maximum voltage across it is dependent on the power source, a large capacitor produces a larger current.

Procedure:

1. Adjust all the controls to the left such that d = 0.4, L = 1.5 and time = 0.
1. Observe that the capacitor discharges almost all its stored charge after about 50 s.
2. The capacitor has a capacitance of $0.481\mu \text{F}$ and its maximum stored charge is $21.6 \mu \text{C}$
2. Increase the separation d to 0.7.
1. Observe that the capacitance is now $0.275\mu \text{F}$ and maximum charge $12.4 \mu \text{C}$. This means that increasing the distance between the plates decreases the capacitance, and correspondingly the charge stored, since $Q=CV$ .
2. Lesser stored charge means that the current stops flowing after about 30 s.
3. Increase the length of side to 3.0.
1. Observe that the capacitance is now $1.1 \mu \text{F}$ and that the charge stored is now $49.4 \mu \text{C}$ . Increasing the cross-sectional area increases the capacitance and the stored charge.
2. Since the stored charge is larger, the capacitor takes a longer time to discharge to 0 V.

### Summary

1. Capacitance depends on the physical property. The capacitance is larger is the distance between the plates is small and the cross-sectional is large.
2. A large capacitor stores a larger charge, and takes a longer time to discharge. Conversely, it should take a larger time to charge a large capacitor to its maximum voltage.

# Capacitance

### What is a capacitor?

A capacitor is an electrical component that stores charge. It is usually made by having two parallel plates thinly separated by an insulating material. This material is also known as a dielectric. Current flows through an insulator if the voltage across it is sufficient. In a capacitor, the conducting plates are close enough for the current to flow, even through it is separated by the insulating dielectric. However, because of the dielectric, some charges are stored on the plates as the current flows, resulting in higher potential difference as time passes. When the potential difference across the capacitor equals to the e.m.f of the circuit, current stops flowing.

### Physics of capacitance

Capacitance is defined as

Capacitance is the ratio of the charge stored to potential difference across a capacitor.

Mathematically,

\begin{aligned}C=\frac{Q}{V} \end{aligned}

The SI  unit of capacitance is farad (F).

A capacitor of 1 farad has a stored charge of 1 C when the potential difference across it is 1 V.

### Combined Capacitance

Capacitors can be connected either in series or parallel.

In a series connection, the combined capacitance can be found from the relationship

\begin{aligned} V_\text{total} &= V_1 + V_2 + ...\\\frac{Q}{C_\text{total}} &= \frac{Q}{C_1}+\frac{Q}{C_2}+...\end{aligned}

Since the charge $Q$ stored in each capacitance is the same,

\begin{aligned}\frac{1}{C_\text{total}} &= \frac{1}{C_1}+\frac{1}{C_2}+...\end{aligned}

In a parallel connection, the combined resistance is derived as follow:

\begin{aligned} Q_\text{total}&=Q_1+Q_2+...\\C_\text{total}V&=C_1V+C_2V+...\\\end{aligned}

Since the potential difference $V$ is the same in a parallel circuit,

\begin{aligned} C_\text{total}=C_1+C_2+... \end{aligned}

## Summary

1. Capacitance is the ratio of the charge stored to the potential difference across the component.
2. SI unit of capacitance is farad (F).
3. The total capacitance of series connected capacitors is \begin{aligned} \frac{1}{C_\text{total}}=\frac{1}{C_1}+\frac{1}{C_2}+... \end{aligned}
4. The total capacitance of parallel connected capacitors is \begin{aligned} C_\text{total} = C_1+C_2+... \end{aligned}

## Review

### Question 1

The figure shows three capacitors connected in series with a cell of e.m.f. 3.0 V.

Calculate the p.d. across each capacitor.

### Solution

Since the charge stored in each capacitor is the same as the charge stored across all the capacitors, we shall go ahead and find the charge stored in the combined capacitors and then use it to calculate the potential difference across each capacitor.

\begin{aligned} \frac{1}{C_\text{total}} &= \frac{1}{100 \times 10^{-3}}+\frac{1}{200 \times10^{-3}} + \frac{1}{400 \times 10^{-3}}\\C_\text{total}&= 0.0571\text{ F}\\ Q_\text{total}&=3.0 \times 0.0571\\&=0.1713\text{ C}\end{aligned}

Hence,

\begin{aligned} V_1&=\frac{0.1713}{100 \times 10^{-3}}\\&=1.713\text{ V}\\V_2&=\frac{0.1713}{200 \times 10^{-3}}\\&=0.857\text{ V} \\V_3&=\frac{0.1713}{400 \times 10^{-3}}\\&=0.428\text{ V}\end{aligned}

You can observe that the sum of all the three p.d. is the e.m.f. of the cell.

# Gravitation and Circular Motion

### Why does a orbiting satellite not fall to the Earth?

A common misconception among students is that orbiting satellites do not experience gravity. It is argued that since these satellites are not falling to the Earth, they must experience no gravity. This thought is not true because the very action of orbiting around the Earth shows that the satellite is under the influence of Earth’s gravity. Then why is the satellite not falling to the Earth?

The below simulation explains why satellites are not falling back to the Earth. In fact, these satellites are actually in the process of falling to the Earth. The simulation explains clearly how this works.

Satellite orbiting the Earth

So the answer to the above question is that the satellite is in fact, falling to the Earth. However, the Earth’s surface curves away at the same rate the the satellite is falling towards Earth. Hence, the satellite never falls onto the Earth.

It is important to note that this can happen because the satellite possesses an initial velocity that is parallel to the Earth’s surface. Without sufficient velocity of this component, the satellite would fall towards Earth.

### Gravitational force and circular motion

Planets revolve around the Sun and the Moon revoles around the Earth. These orbits are generally approximately circular, and the centripetal force to pull them around the orbits is the gravitational force.

\begin{aligned} \frac{GMm}{r^2} &= \frac{mv^2}{r} \\ \frac{GM}{r} &= {v^2}\\v&=\sqrt{\frac{GM}{r}}\end{aligned}

We can see that the linear speed, hence the angular speed, of an orbiting body is independent of the mass of the body. It is only dependent on the mass of the body producing the gravitational field and the distance of the orbiting body from the centre of the main mass.

To find the relationship between the radius and the period of orbit,

\begin{aligned} (\frac{2\pi r}{T})^2 &= \frac{GM}{r}\\T^2 &= \frac{4\pi^2}{GM}r^3 \end{aligned}

Hence we can see that the square of the period is proportional to the cube of the radius of orbit. The is the Kepler’s third law.

### Summary

1. When an object orbits around another mass, the centripetal force is gravitational in nature.

2. The square of the period is proportional to the cube of the radius of orbit. This is also known as the Kepler’s third law.

### Review

#### Question 1

Two stars orbit each other in a time of 3.0 years. Calculate the angular speed \begin{aligned} \omega \end{aligned} for each star. Given that the ratio \begin{aligned} \frac{M_1}{M_2}=2.0 \end{aligned} , and the separation of the starts is $3.0 \times 10^{12} \text{ m}$ , calculate the radii \begin{aligned} R_1 \end{aligned} and \begin{aligned} R_2 \end{aligned} .

#### Solution:

Angular speed is the angular displacement per unit time. Hence,

\begin{aligned}\omega &= \frac{2 \pi}{3.0 \times 365 \times 24 \times 3600}\\&=6.6\times 10^{-8}\text{rad s}^{-1} \end{aligned}

The two stars must be orbiting with the same angular speed. Otherwise, there will be a point in time when the inner star catches up and the centre of mass would change position. The gravitational force between the two stars would then change, changing the centripetal force. This is not possible for a stable star system, hence the angular speeds of the two stars must be the same.

Equating the gravitational force to the centripetal force,

\begin{aligned} \frac{GM_1M_2}{(R_1 + R_2)^2 } &= M_2 R_2 \omega^2\\\frac{GM_1}{(R_1 + R_2)^2} &= R_2 \omega^2 \end{aligned}

Similarly,

\begin{aligned} \frac{GM_2}{(R_1 + R_2)^2} &= R_1 \omega^2 \\\frac{M_1}{M_2} &= \frac{R_2}{R_1}\end{aligned}

Since \begin{aligned} \frac{M_1}{M_2} = 2.0 \end{aligned} ,

\begin{aligned} \frac{R_2}{R_1}&=2.0\\R_2&=2.0R_1\\R_1+2R_1 &= 3.0 \times 10^{12}\\R_1&=1.0\times 10^{12}\text{ m}\\R_2 &= 2.0 \times 10^{12} \text{ m} \end{aligned}