Energy concepts

4 Energy Concepts Learners should be able to demonstrate and apply their knowledge and understanding of:
a)	The idea that work is the product of a force and distance moved in the direction of the force when the force is constant
b)	The calculation of the work done for constant forces, when the force is not along the line of motion([math] \text{Work done} = Fx cos⁡θ[/math])
c)	The principle of conservation of energy including knowledge of gravitational potential energy(mg∆h), elastic potential energy ([math]\frac{1}{2} kx^2 [/math]) and kinetic energy ([math]\frac{1}{2} mv^2 [/math])
d)	The work-energy relationship: [math]Fx = \frac{1}{2} mv^2 – \frac{1}{2} mu^2 [/math]
e)	Power being the rate of energy transfer
f)	Dissipative forces for example, friction and drag cause energy to be transferred from a system and reduce the overall efficiency of the system
g)	[math]\text{Efficiency} = \left( \frac{\text{Useful energy transfer}}{\text{Total energy input}} \right) \times 100\% [/math]

Learners should be able to demonstrate and apply their knowledge and understanding of:
a) Work as a Product of Force and Distance
⇒ Definition of Work:
Work is done when a force is applied to an object, and the object moves in the direction of the force.
For a constant force, the work W is given by:
[math]W = Fdcosθ[/math]
Where:
– W: Work done (in joules, J).
– F: Force applied (in newtons, N).
– d: Displacement of the object (in meters, m).
– θ: The angle between the force and the displacement.
Figure 1 Work done
If θ = 0°: The force is in the same direction as the motion, and cosθ =1 , so
[math]W = Fd[/math]
If θ = 90°: The force is perpendicular to the motion, and cosθ = 0, so W=0 (no work is done).
b) Work Done for Constant Forces When the Force is Not Along the Line of Motion
When a force is applied at an angle to the direction of motion:
[math] W = Fdcosθ [/math]
Example:
– Consider pulling a sled at an angle θ to the ground.
– The force F has two components: horizontal [math]F_x = Fcos θ [/math] and vertical [math]F_y = Fcin θ [/math]
– Only the horizontal component contributes to work done in moving the sled.
Figure 2 Work done by a constant force
c) The Principle of Conservation of Energy
⇒ Definition:
Energy cannot be created or destroyed, only transferred or transformed from one form to another. The total energy in a closed system remains constant.
⇒ Types of Energy and Their Formulas:
1. Gravitational Potential Energy (GPE):
[math]E_g = \frac{1}{2} mv^2 [/math]
– [math]E_g [/math]: Gravitational potential energy (in joules, J).
– m: Mass of the object (in kilograms, kg).
– g: Acceleration due to gravity (9.8 m/s²)
– Δh: Change in height (in meters, m).
2. Kinetic Energy (KE):
[math]E_k = \frac{1}{2} mv^2 [/math]
– [math]E_k [/math]: Kinetic energy (in joules, J).
– m: Mass of the object (in kilograms, kg).
– v: Velocity of the object (in meters per second, m/s).
3. Elastic Potential Energy:
[math]E_e = \frac{1}{2} kx^2 [/math]
– [math]E_e [/math]: Elastic potential energy (in joules, J).
– k: Spring constant (in newtons per meter, N/m).
– x: Extension or compression of the spring (in meters, m).
Figure 3 Principle of conservation of energy

d) Work-Energy Relationship:
⇒ Definition:
The work done on an object result in a change in its kinetic energy. This is known as the work-energy theorem:
[math]W = ∆KE = \frac{1}{2} kv^2 – \frac{1}{2} ku^2 [/math]
Where:
– v: Final velocity.
– u: Initial velocity.
[math]W = FΔx[/math]
If the force is along the displacement.
⇒ Explanation of Energy Conservation with Work-Energy Principle
When work is done on an object:
– If the object gains speed, its kinetic energy increases.
– If the object is lifted, its gravitational potential energy increases.
– If the object compresses or stretches a spring, its elastic potential energy increases.
Example:
– A car accelerates due to the work done by the engine. The work translates to an increase in the car’s kinetic energy:
[math]W = Fd \cos \theta \\ W = \frac{1}{2} k v^2 – \frac{1}{2} k u^2[/math]
⇒ Example Problems
Work Done with Force at an Angle:
– A box is dragged across the floor by a force of 50 N at an angle of 30° to the horizontal over a distance of 10 m.
[math]W = Fd \cos \theta \\
W = 50 \cdot 10 \cdot \cos 30^\circ \\
W = 433 \, \text{J} [/math]
Conservation of Energy:
– A ball of mass 2 kg is dropped from a height of 5 m.
– Initial GPE:
[math]W = mgh \\
W = (2)(9.8)(5) \\
W = 98 \, \text{J} [/math]
– At the ground, the GPE is converted to KE:
[math]KE = \frac{1}{2} mv^2 \\
98 = \frac{1}{2}(2)v^2 \\
v = \sqrt{98} \\
v = 9.9 \, \text{m/s} [/math]
Elastic Potential Energy:
– A spring with k=100 N/m is compressed by 0.2 m. The energy stored:
[math]E_e = \frac{1}{2} kx^2 \\
E_e = \frac{1}{2}(100)(0.2)^2 \\
E_e = 2 \, \text{J} [/math]
Figure 4 Conservation of energy
e) Power as the Rate of Energy Transfer
⇒ Definition of Power:
Power is the rate at which energy is transferred or work is done.
Mathematically, power P is defined as:
[math]P = \frac{E}{t}[/math]
Where:
– P: Power (in watts, W).
– E: Energy transferred or work done (in joules, J).
– t: Time taken (in seconds, s).
⇒ Alternate Form Using Force and Velocity:
When a force F is applied to an object moving at a velocity v, power can be expressed as:
[math]P = Fv [/math]
– This equation is valid only when F and v are in the same direction.
Power is measured in watts (W), whereW = 1J/s.
High power means a faster rate of energy transfer.
Example:
– A motor does 600 J of work in 10 seconds. The power output is:
[math]P = \frac{E}{t} \\
P = \frac{600}{10} \\
P = 60 \, \text{W} [/math]

f) Dissipative Forces
⇒ Definition:
Dissipative forces, such as friction and drag, resist motion and cause energy to be transferred out of a system, usually as thermal energy.
Figure 5 Dissipative Forces
⇒ Examples of Dissipative Forces:
1. Friction:
– Acts between surfaces in contact and opposes relative motion.
– Energy is lost as heat, reducing the efficiency of machines or systems.
2. Drag:
– A resistive force acting on an object moving through a fluid (e.g., air or water).
– Energy is dissipated as heat due to the interaction with the fluid.
⇒ Effects of Dissipative Forces:
Dissipative forces reduce the total mechanical energy of a system.
In real-life scenarios, they decrease the efficiency of systems by converting useful energy into heat or sound.
⇒ Example:
A car engine generates energy to move the vehicle forward, but some energy is lost due to friction in the engine and drag acting on the car.
g) Efficiency
⇒ Definition:
Efficiency is a measure of how well a system converts input energy into useful output energy.
⇒ Efficiency Formula:
[math]\text{Efficiency} = \left( \frac{\text{Useful energy transfer}}{\text{Total energy input}} \right) \times 100\% [/math]
– Efficiency is expressed as a percentage.
– A perfectly efficient system (100%) converts all input energy into useful energy, but in practice, some energy is always lost due to dissipative forces.
⇒ Factors Affecting Efficiency:
1. Friction:
– Causes loss of energy as heat.
2. Air resistance:
– Dissipates energy into the surroundings.
3. Electrical resistance:
– Generates heat in electrical systems.
4. Sound energy:
– Unintended loss in systems with moving parts.
⇒ Practical Examples
1. Light Bulb:
Input: Electrical energy.
Useful output: Light energy.
Loss: Heat energy.
Efficiency:
[math]\text{Efficiency} = \left( \frac{\text{Light energy output}}{\text{Electrical energy input}} \right) \times 100\%[/math]
Figure 6 Light bulb efficiency
2. Car Engine:
Input: Chemical energy in fuel.
Useful output: Kinetic energy to move the car.
Loss: Heat, sound, and friction.
Efficiency:
[math]\text{Efficiency} = \left( \frac{\text{Kinetic energy output}}{\text{Chemical energy input}} \right) \times 100\% [/math]
Figure 7 Car engine efficiency
3. Electric Motor:
Input: Electrical energy.
Useful output: Mechanical energy to rotate a shaft.
Loss: Heat and sound.

Figure 8 Electric Motor
⇒ Improving Efficiency:
Reduce friction by using lubricants.
Streamline objects to reduce air resistance.
Use materials with low electrical resistance.
⇒ Worked Example
Problem:
– A motor uses 2000 J of electrical energy to lift a 150 kg mass through a height of 1 m. Calculate:
1. The useful energy output.
2. The efficiency of the motor.
Solution:
1. Useful Energy Output:
– Gravitational potential energy:
[math]E_p = mgh \\
E_p = (150)(9.8)(1) \\
E_p = 1470 \, \text{J} [/math]
2. Efficiency:
[math]\text{Efficiency} = \left( \frac{\text{Useful energy transfer}}{\text{Total energy input}} \right) \times 100\% \\
\text{Efficiency} = \frac{1470}{2000} \times 100\% \\
\text{Efficiency} = 73.5\%[/math]