1. Work:

• Work is defined as the transfer of energy from one object to another through a force applied over a displacement.
• In other words, work is done when a force is applied to an object and the object moves in the direction of the force.
• Formula:
• Work is a scalar quantity, meaning it has only magnitude (amount of work) but no direction.
• The unit of work is typically measured in Joules (J).
• Conditions:
  • Positive work: when the force and displacement are in the same direction.
  • W = Fdcosθ         θ=0°
    W = Fd(1)
    W = Fd
  • 
  • Figure 1 Displacement covers in the direction of applied force

• 
  Figure 2 negative work

• Negative work: when the force and displacement are in opposite directions (e.g., lowering a book)

  W = Fdcosθ  θ = 180°
  W = Fd(-1)
  W = – Fd

• 
  Figure 3 Zero work done

  Zero work: when the force and displacement are perpendicular (e.g., pushing a wall)

• W=Fdcosθ θ = 90
  W=Fd(0)
  W=0

⇒ Examples

(1)

• Tracy is doing some weight training.
• She is holding two weights but she is not moving them.
• She gets tired holding them out, and her arms convert chemical energy into heat energy, but no work is done as the weights do not move.
• Tracy does work when she lifts the weights.

Figure 4 Tracy is using energy but not working.

(2)

• Three people are helping to push-start a friend’s car.
• Salim and Anne are pushing at the back of the car.
• Here Salim and Anne are working, because they are pushing along the direction in which the car is travelling.
• Jim does no work as he is pushing at right angles to the direction of motion.

Figure 5 Salim and Anne are working but Jim is not working as he is pushing at right angles to the car

(3)

• Calculate the work done from Figure 6 when the object is displaced 40 m from its original position.

Figure 6 Calculate the work done by considering a small displacement,[math]\vec{d}[/math] , when the applied force is [math]\vec{F}[/math] 

2. Calculating the energy

• Work is converted to heat energy when it is done against resistive forces, like moving a luggage through an airport.
• As a result, the suitcase’s wheels and the area around them somewhat heated.
• Energy may also be transformed into many forms via work.
• This concept allows us to obtain the formulae for kinetic energy, elastic potential energy, and gravitational potential energy.

⇒ Gravitational potential energy

• Gravitational potential energy is the energy an object possesses due to its height or position in a gravitational field.

  
  Figure 7 gravitational potential energy converted into the kinetic energy

• It’s the energy an object has because of its potential to fall or move downward.
• Gravitational potential energy is a type of potential energy, which means it has the potential to become kinetic energy (the energy of motion).
• The higher an object is, the more gravitational potential energy it has.
• Gravitational potential energy is always relative to a reference level, such as the ground or a table.
• When an object falls or moves downward, its gravitational potential energy is converted into kinetic energy.
• A ball at the top of a point has gravitational potential energy due to its height.
• We can calculate the gravitational potential energy by using formula
• ∆E_p = mg∆h
• Where
  • ΔE_p is gravitational potential energy
  • m is mass of an object
  • g is gravitational acceleration
  • ∆h is height of an object in which potential energy store
• In Figure 8 a load with a weight W, has been lifted through a height ∆h.
  The work done = increase in gravitational potential energy.
• Work = W * Δh

Weight W=mg

• So, the increase in potential energy,  , is given by
• ∆E_p = mg∆h
• 
• Figure 8 A load with a weight W lifted through a height Δh.

⇒ Elastic potential energy

• Elastic potential energy is the energy stored in an object that has been stretched, compressed, or deformed in some way.
• It’s the energy an object has due to its elastic properties, like a spring or a rubber band.
• When an object is stretched or compressed, its molecules are moved apart or pushed together, creating a force that wants to return the object to its original shape. This force is called the restoring force, and the energy associated with it is the elastic potential energy.

  
  Figure 9 a spring may be stretched

• The more an object is stretched or compressed, the more elastic potential energy it has.
• When an object is released from its stretched or compressed state, the elastic potential energy is converted into kinetic energy.
• Elastic potential energy (E_ep) is stored in a stretched spring.
• It has the ability to transform its stored energy into the kinetic energy when it is released.
• The relationship between the force needed to stretch the band and its pullback distance is seen in Figure 10. We must compute the stored energy using the average force since the force varies.
• [math]\Delta E_{\text{ep}} = \vec{F}_{\text{av}} * \vec{s}[/math]
• In the case of Figure 10, where the force to stretch the band is proportional to the distance moved.
• [math]\text{Average force} = \frac{1}{2} \, \text{final force}[/math]
• So,         [math]\Delta E_{\text{ep}} = \frac{1}{2} F s[/math]
• The stored elastic potential energy can also be calculated more generally using the area under the force-extension graph.

Figure 10 How the force required to stretch the band depends on the distance it is pulled back.

⇒ Kinetic energy

• Kinetic energy is the energy of motion. It’s the energy an object possesses due to its motion, and it’s defined as the work required to bring the object to rest.
• In other words, it’s the energy an object has because it’s moving.
• Kinetic energy is a scalar quantity, meaning it has only magnitude (amount of energy) but no direction.
• The more massive an object is and the faster it moves, the more kinetic energy it has.
• Kinetic energy can be converted into other forms of energy, such as potential energy, and vice versa.
• 
  Figure 11 A constant force accelerates a car
• In Figure 11, a constant force F accelerates a car, starting at rest, over a distance s.
• Work is done to increase the kinetic energy of the car.
• We can use this idea to find a formula for kinetic energy, E_k, in terms of the car’s speed and mass.
• [math]\Delta E_k = F s \quad (1)[/math]
• but from Newton’s Second Law:
• F = ma
• and from the equations of motion:
• [math] s = \frac{1}{2} a t^2[/math]
• Put in equation 1
• [math]\Delta E_k = ma * \frac{1}{2} at^2 \\ \Delta E_k = \frac{1}{2}ma^2t^2 [/math]
• Since
• 
• So the kinetic energy of a body of a mass m, moving at a velocity v, is given by:

   [math] E_k = \frac{1}{2} mv^2 [/math]

  Note this is a scalar quantity because  has no direction

3.The principle of conservation of energy:

• The principle of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another.
• In other words, the total energy of an isolated system remains constant over time.
• This means that the sum of all the different types of energy within a system (kinetic, potential, thermal, electrical, etc.) remains the same at all times.
• Energy can be converted from one form to another, but the total energy remains constant.
• [math]Total kinetic energy = total potential energy[/math]

⇒ Examples:

(1)

• A boy throws a ball upwards with a speed of 16m/s.
• It leaves his hand at a height of 1.5m above the ground.
• Calculate the maximum height to which it rises
  Solution:
• • The ball gains gravitational potential energy as it rises and loses kinetic energy when it leaves the boy’s grasp.
  • The ball stops travelling at its highest point, converting all of its kinetic energy to potential energy.
  • Given data:
    Velocity = 16m/s
  • Find data:
    Maximum height = ?
  • Formula:
    
  • Solution:
    
  • but the total height gained is 13m + 1.5m = 14.5m.

(2)

• A stretched catapult stores 0.7J of elastic potential.
• The catapult is used to launch a marble of mass 0.01kg.
• Calculate the initial speed of the marble.

Given data:
The stored elastic potential energy in the catapult is transferred to the marble’s kinetic energy.
Elastic potential energy ([math]\Delta E_ep)= 0.7J
Mass = 0.01kg
Find data:
Speed (velocity) = V= ?
Formula:

[math] \Delta E_{\text{ep}} = \Delta E_k[/math]

Solution:

Putting Values:

4. Efficiency

• Efficiency is a measure of how well a system or process uses energy to produce a desired output.
• It’s defined as the ratio of useful output energy to input energy.
• Formula:
• [math]\text{Efficiency} = \frac{\text{useful output energy}}{\text{input energy}} \\ \text{Efficiency} = \frac{\text{useful output energy}}{\text{input energy}} \times 100\% [/math]
• Efficiency is usually expressed as a percentage, with 100% being the maximum possible efficiency.
• In reality, no system can achieve 100% efficiency, as some energy is always lost as waste heat, friction, or other forms of energy.
• Types of efficiency:
  1. Mechanical efficiency: measures the efficiency of machines or mechanisms.
  2. Thermal efficiency: measures the efficiency of energy conversion from heat to work.
  3. Electrical efficiency: measures the efficiency of electrical systems or devices.
  4. Energy efficiency: measures the efficiency of energy use in buildings, industries, or processes.
• Efficiency is important because it:
  1. Saves energy and resources
  2. Reduces waste and pollution
  3. Increases productivity and performance
  4. Saves money and reduces costs
• Sankey diagram: A particular flow diagram where the width of the indicated arrows corresponds to the flow amount. This chapter uses Sankey diagrams to illustrate the flow of energy through different processes.
• 

• Figure 12 Sankey diagram

⇒ Example

• A Sankey diagram illustrating the conversion of 100J of electrical energy into different forms of energy may be seen in Figure 12.
• Just 30J of the energy is converted into useable gravitational potential energy.
• The rest is converted to heat and sound energy.
• [math]\text{Efficiency} = \frac{\text{useful output energy}}{\text{input energy}} \\
  \text{Efficiency} = \frac{30 \, \text{J}}{100 \, \text{J}} \\
  \text{Efficiency} = 0.3 \text{ or } 30\% [/math]

5. Power

• Power is the rate at which work is done or energy is transferred.
• It’s the amount of energy transferred per unit time.
• In other words, power is the measure of how quickly energy is used or produced.
• Formula:
• [math] \text{Power} = \frac{\text{energy transferred}}{\text{time}} \quad \text{or} \quad \frac{\text{work done}}{\text{time}} [/math]
• The unit of power is the watt (W), or J/s.
• This definition can be used to produce a useful formula to calculate the power transferred by moving vehicles.
• [math] \text{Power} = \frac{\text{work done}}{\text{time}} \quad \text{or} \quad \frac{\text{W}}{\text{t}} \\ \text{Power} = \frac{\vec{F} \cdot \vec{s}}{t}[/math]

• or

• [math]\text{Power} = \vec{F} \cdot \vec{v} [/math]
• Power (P) is the rate of energy transfer (measured in watts, W)
• Work (W) is the energy transferred (measured in joules, J)
• Time (t) is the time over which the energy is transferred (measured in seconds, s)

⇒ Example:

• A car is moving at a constant speed of 18m/s. The frictional forces acting against the car are 800N in total. The car has a mass of 1200kg.
• a) Calculate the power transferred by the car on a level road.
• b) The car maintains its constant speed while climbing a hill of vertical height 30m in 16s. Calculate the power transferred by the car now.

Given data:

Speed = 18m/s

Force = 800N

Mass = 1200 kg

Find data:
(a) Power transferred by the car on a level road = ?
(b) Power transferred by the car when car climbing a hill = ?

Vertical height = 30m

Time taken = 16s

Formula:

[math]\text{(a)} \quad \text{Power} = \vec{F} \cdot \vec{v}\\
\text{(b)} \quad \text{Power} = \text{previous power} + \frac{W}{t}\\
\text{Power} = \text{previous power} + \frac{mg \Delta h}{t} [/math]

Solution:

[math] \text{(a)} \quad \text{Power} = \vec{F} \cdot \vec{v}\\
\text {Power} = 800 * 18 [/math]

(a) Power transferred by the car on a level road

[math] \text{Power} = 14 \, \text{KW} [/math]

(b) Power transferred by the car when car climbing a hill

[math]\text{Power} = \text{previous power} + \frac{mg \Delta h}{t} \\
\text{Power} = 14000 + \frac{1200(9.8)(30)}{16} \\
\text{Power} = 14000 + 22050\\[/math]
[math]\text{Power} = 36 \text{KW}[/math]