Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Mechanical Energy
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Mechanical Energy
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Subject: Physics
Class: Senior Secondary 2
Term: 1st Term
Week: 6
Theme: Conservation Principles
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Students shouldbe able to:Calculate the kinetic and potentialenergy of abody. Verifyconservationof energyprinciplesand showthat the to talenergy is conserved fora given set of data on the energy of aparticle in aconservationfield. Determine amachine and list at leastfive simplemachines.
Define and calculate:(A) For ce ratio(B) Velocity ratio(C) Effiency for as implemachine and write the mathematicalrelationshipbetween (a),(b) and (c). State howfriction can bereduced in the moving partsof a givenmachine. Perform as impleexperimentusing a springbalance to determine the co-efficient of frictionbetween twosurface.
2. 1. Mechanical Energy Mechanical energy is the energy an object possesses due to its motion or its position. It is the sum of kinetic energy and potential energy. 2.1.
1. Kinetic Energy (KE)
Definition: Kinetic energy is the energy possessed by an object due to its motion. Any object that is moving has kinetic energy.
Formula: $KE = \frac{1}{2}mv^2$ Where: $m$ = mass of the object (in kilograms, kg) $v$ = speed or velocity of the object (in metres per second, m/s)
Unit: The SI unit for kinetic energy is the Joule (J).
Explanation: The formula shows that kinetic energy is directly proportional to the mass of the object and the square of its velocity. This means a heavier object moving at the same speed has more KE, and an object moving faster has significantly more KE (due to the squaring of velocity). Nigerian Context
Example: A moving "okada" (motorcycle taxi) possesses kinetic energy. A faster okada or one carrying more passengers (higher mass) will have more kinetic energy. 2.1.
2. Potential Energy (PE)
Definition: Potential energy is the energy stored in an object due to its position or state.
Types: Gravitational Potential Energy (GPE): Energy stored in an object due to its height above a reference point (usually the ground).
Formula: $GPE = mgh$ Where: $m$ = mass of the object (in kilograms, kg) $g$ = acceleration due to gravity (approximately $9.8 \text{ m/s}^2$ or $10 \text{ m/s}^2$ for simplicity in calculations) $h$ = height of the object above the reference point (in metres, m)
Unit: The SI unit for gravitational potential energy is the Joule (J).
Explanation: GPE depends on the mass of the object, the strength of gravity, and how high the object is lifted. The higher an object, the more GPE it stores. Nigerian Context
Example: Water stored behind the dam wall at Kainji has significant gravitational potential energy, which is converted to electrical energy. A bag of rice lifted onto a market stall has GP
E. Elastic Potential Energy (EPE): Energy stored in an elastic material (like a spring or stretched rubber band) when it is stretched or compressed. While mentioned for completeness, the focus for SS2 mechanical energy typically emphasizes GPE. 2.
2. Conservation of Mechanical Energy Principle: The Law of Conservation of Energy states that energy cannot be created or destroyed, but can only be transformed from one form to another. In the context of mechanical energy, this means that in an isolated system where only conservative forces (like gravity) are doing work, the total mechanical energy (KE + PE) remains constant.
Mathematical Relationship: Total Mechanical Energy ($ME$) = $KE + PE$ In a conservative field, $KE_1 + PE_1 = KE_2 + PE_2$ (where 1 and 2 refer to initial and final states).
Explanation: As an object falls, its height ($h$) decreases, so its GPE decreases. Simultaneously, its velocity ($v$) increases, so its KE increases. The decrease in GPE is exactly equal to the increase in KE, meaning their sum remains constant.
Example: A stone falling from a height. At its highest point (rest), $KE=0$, $ME = GPE_{max}$. As it falls, GPE decreases, KE increases. Just before hitting the ground, $GPE=0$, $ME = KE_{max}$. Throughout the fall (ignoring air resistance), $GPE + KE = \text{Constant}$.
Worked Example 1: Calculation of KE and PE A 2 kg coconut falls from a tree at a height of 15 m. (a) Calculate its potential energy at the height of 15 m. (b) Calculate its kinetic energy just before it hits the ground (assume $g = 10 \text{ m/s}^2$).
Solution: (a)
Potential Energy at 15 m: Given: $m = 2 \text{ kg}$, $h = 15 \text{ m}$, $g = 10 \text{ m/s}^2$ $GPE = mgh = 2 \text{ kg} \times 10 \text{ m/s}^2 \times 15 \text{ m} = 300 \text{ J}$ The potential energy of the coconut at 15 m is 300 J. (b) Kinetic Energy just before hitting the ground: According to the principle of conservation of mechanical energy, the initial potential energy is converted entirely even some Nigerian tricycles "keke NAPEP" are designed with some aerodynamic consideration).
5. Using Air Cushioning: Creating a layer of air between surfaces to eliminate contact (e.g., hovercrafts). Less common in everyday Nigerian machinery. 2.
6. Coefficient of Friction ($\mu$)
Definition: A dimensionless quantity that represents the ratio of the force of friction between two surfaces to the normal force pressing the surfaces together. It indicates the "stickiness" or "slipperiness" of the surfaces.
Formula: $\mu = \frac{F_f}{N}$ Where: $F_f$ = Frictional force (in Newtons, N) $N$ = Normal reaction force (the force perpendicular to the surface, equal to the weight of the object on a horizontal surface, $N=mg$) (in Newtons, N) Experimental Determination (using a spring balance): Setup: A block of known mass ($m$) is placed on a horizontal surface. A spring balance is attached to the block.
Procedure:
1. Measure the mass ($m$) of the block. Calculate its weight ($N = mg$).
2. Pull the block horizontally using the spring balance at a constant, slow speed.
3. Read the force ($F_f$) indicated by the spring balance. This force represents the kinetic frictional force.
4. Repeat the experiment several times and take an average reading for $F_f$.
5. Calculate the coefficient of kinetic friction using $\mu_k = \frac{F_f}{N}$.
For static friction: Pull the block until it just begins to move. The maximum reading before movement gives the maximum static friction ($F_{s,max}$). Then $\mu_s = \frac{F_{s,max}}{N}$. Nigerian Context
Example: Determining the coefficient of friction between a wooden crate and a concrete floor, or a rubber shoe sole and wet tarmac. 1, it multiplies distance or speed (e.g., fishing rod). If MA = 1, it only changes the direction of force (e.g., single fixed pulley). MA has no unit as it is a ratio of forces. 2.4.
2. Velocity Ratio (VR)
Definition: The ratio of the distance moved by the effort to the distance moved by the load. It is a theoretical value determined by the geometry of the machine.
Formula: $VR = \frac{\text{Distance moved by Effort}}{\text{Distance moved by Load}} = \frac{d_E}{d_L}$ Explanation: VR has no unit. It indicates the speed advantage the machine provides. For ideal machines, VR is often a fixed value based on construction (e.g., for a block and tackle pulley system, VR is the number of ropes supporting the load). 2.4.
3. Efficiency ($\eta$)
Definition: The ratio of work output to work input, usually expressed as a percentage. It indicates how effectively the machine converts input energy into useful output work.
Formula: $\eta = \frac{\text{Work Output}}{\text{Work Input}} \times 100\%$ $\eta = \frac{L \times d_L}{E \times d_E} \times 100\%$ Relationship between MA, VR, and Efficiency: Since $W_{out} = L \times d_L$ and $W_{in} = E \times d_E$, we can write: $\eta = \frac{L/E}{d_E/d_L} \times 100\%$ Therefore, $\eta = \frac{MA}{VR} \times 100\%$ Explanation: No machine is 100% efficient due to energy losses, primarily through friction. The lost energy is usually converted into heat and sound.
Worked Example 2: Machine Calculations A wheelbarrow is used to carry a load of 150 N. An effort of 50 N is applied to lift the wheelbarrow. If the load moves up by 0.1 m when the effort moves down by 0.5 m, calculate: (a) The Mechanical Advantage (MA). (b) The Velocity Ratio (VR). (c) The Efficiency ($\eta$) of the wheelbarrow.
Solution: Given: Load ($L$) = 150 N, Effort ($E$) = 50 N, Distance moved by load ($d_L$) = 0.1 m, Distance moved by effort ($d_E$) = 0.5 m. (a)
Mechanical Advantage (MA): $MA = \frac{L}{E} = \frac{150 \text{ N}}{50 \text{ N}} = 3$ The Mechanical Advantage is 3. (b)
Velocity Ratio (VR): $VR = \frac{d_E}{d_L} = \frac{0.5 \text{ m}}{0.1 \text{ m}} = 5$ The Velocity Ratio is 5. (c) Efficiency ($\eta$): $\eta = \frac{MA}{VR} \times 100\% = \frac{3}{5} \times 100\% = 0.6 \times 100\% = 60\%$ Alternatively: $W_{out} = L \times d_L = 150 \text{ N} \times 0.1 \text{ m} = 15 \text{ J}$ $W_{in} = E \times d_E = 50 \text{ N} \times 0.5 \text{ m} = 25 \text{ J}$ $\eta = \frac{W_{out}}{W_{in}} \times 100\% = \frac{15 \text{ J}}{25 \text{ J}} \times 100\% = 0.6 \times 100\% = 60\%$ The efficiency of the wheelbarrow is 60%. 2.
5. Friction Definition: Friction is a force that opposes the relative motion or tendency of motion between two surfaces in contact.
Effects: Friction can be both useful (e.g., allowing cars to move, enabling us to walk) and undesirable (e.g., causing wear and tear in machines, reducing efficiency).
Ways to Reduce Friction in Machines:
1. Lubrication: Applying oil, grease, or graphite between moving surfaces to create a thin layer that reduces direct contact (e.g., lubricating engine parts of a car, applying grease to bicycle chains).
2. Using Ball Bearings or Roller Bearings: These convert sliding friction into rolling friction, which is much smaller. They are used in axles of vehicles, bicycle wheels, and industrial machinery.
3. Polishing Surfaces: Making surfaces smoother reduces the interlocking of irregularities and thus reduces friction (e.g., smooth gears).
4. Streamlining: Shaping objects to reduce air or fluid resistance (a type of friction) (e.g., shape of cars, airplanes, boats, and even some Nigerian tricycles "keke NAPEP" are designed with some aerodynamic consideration).
5. Using Air Cushioning: Creating a layer of air between surfaces to eliminate contact (e.g., hovercrafts). Less common in everyday Nigerian machinery. 2.
6. Coefficient of Friction ($\mu$)
Definition: A dimensionless quantity that represents the ratio of the force of friction between two surfaces to the normal force pressing the surfaces together. It indicates the "stickiness" or "slipperiness" of the surfaces.
Formula: $\mu = \frac{F_f}{N}$ Where: $F_f$ = Frictional force (in Newtons, N) $N$ = Normal kinetic energy just before it hits the ground (assume $g = 10 \text{ m/s}^2$).
Solution: (a)
Potential Energy at 15 m: Given: $m = 2 \text{ kg}$, $h = 15 \text{ m}$, $g = 10 \text{ m/s}^2$ $GPE = mgh = 2 \text{ kg} \times 10 \text{ m/s}^2 \times 15 \text{ m} = 300 \text{ J}$ The potential energy of the coconut at 15 m is 300 J. (b) Kinetic Energy just before hitting the ground: According to the principle of conservation of mechanical energy, the initial potential energy is converted entirely into kinetic energy just before hitting the ground (ignoring air resistance). Initial total mechanical energy ($ME_{initial}$) = $GPE_{initial} + KE_{initial}$ At 15 m height, $KE_{initial} = 0$ (coconut starts from rest). So, $ME_{initial} = 300 \text{ J} + 0 = 300 \text{ J}$. Final total mechanical energy ($ME_{final}$) = $GPE_{final} + KE_{final}$ Just before hitting the ground, $GPE_{final} = 0$ (height is 0 m). So, $ME_{final} = 0 + KE_{final}$.
By conservation of mechanical energy: $ME_{initial} = ME_{final}$ $300 \text{ J} = KE_{final}$ The kinetic energy of the coconut just before it hits the ground is 300 J. (Alternatively, one could calculate velocity using $v^2 = u^2 + 2gs$ and then $KE = \frac{1}{2}mv^2$, but conservation of energy is more direct here.) 2.
3. Machines Definition: A machine is a device that helps us to do work more easily or more quickly, or to change the direction of a force. It does not create energy but transforms or transmits it.
Simple Machines: These are the basic mechanical devices for applying force and doing work.
1. Lever: A rigid bar that pivots around a fixed point called a fulcrum (e.g., see-saw, crowbar, bottle opener, wheelbarrow, human arm).
2. Pulley: A wheel with a grooved rim over which a rope or cable passes, used to change the direction of a force or to gain mechanical advantage (e.g., flagpole, well bucket system, construction cranes).
3. Inclined Plane: A flat surface set at an angle to the horizontal, used to move objects to a higher or lower level with less force (e.g., ramp, slide, winding road up a hill).
4. Wheel and Axle: A wheel attached to a smaller rod (axle) so that both rotate together (e.g., car steering wheel, bicycle wheels, door knob, grinding stone).
5. Wedge: Two inclined planes joined back-to-back, used to split objects or hold them in place (e.g., axe, knife, chisel, doorstop).
6. Screw: An inclined plane wrapped around a cylinder, used to fasten objects or lift heavy loads (e.g., wood screw, car jack, vice). 2.
4. Terms Related to Simple Machines Effort (E): The force applied to the machine by the user. (Unit: Newton, N)
Load (L): The resistance or force overcome by the machine. (Unit: Newton, N) Distance moved by Effort ($d_E$): The distance through which the effort moves. (Unit: metre, m) Distance moved by Load ($d_L$): The distance through which the load moves. (Unit: metre, m) Work Input ($W_{in}$): The work done on the machine by the effort. $W_{in} = E \times d_E$. (Unit: Joule, J) Work Output ($W_{out}$): The work done by the machine on the load. $W_{out} = L \times d_L$. (Unit: Joule, J) 2.4.
1. Force Ratio (Mechanical Advantage, MA)
Definition: The ratio of the load (force overcome by the machine) to the effort (force applied to the machine). It indicates how many times the machine multiplies the applied force.
Formula: $MA = \frac{\text{Load}}{\text{Effort}} = \frac{L}{E}$ Explanation: A machine with MA > 1 is a force multiplier (e.g., car jack). If MA < 1, it multiplies distance or speed (e.g., fishing rod). If MA = 1, it only changes the direction of force (e.g., single fixed pulley). MA has no unit as it is a ratio of forces. 2.4.
2. Velocity Ratio (VR)
Definition: The ratio of the distance moved by the effort to the distance moved by the load. It is a theoretical value determined by the geometry of the machine.
Formula: $VR = \frac{\text{Distance moved by Effort}}{\text{Distance moved by Load}} = \frac{d_E}{d_L}$ Explanation: VR has no unit. It indicates the speed
Hydroelectric Power Generation (e.g., Kainji Dam): The concept of gravitational potential energy is fundamental to hydroelectric power. Water stored at a high elevation possesses GPE, which is converted to kinetic energy as it flows down, then to mechanical energy (turning turbines), and finally to electrical energy. This directly relates to Nigeria's energy infrastructure.
Transportation and Road Safety: Understanding kinetic energy is crucial for vehicle design and road safety. The KE of a moving vehicle determines the impact force in a collision. Braking systems in cars, buses, and motorcycles rely on converting kinetic energy into heat energy through friction. This informs safe driving practices and the need for well-maintained vehicles. Traditional Tools and Agricultural Machinery: Many traditional Nigerian tools like hoes, cutlasses, grinding stones (millstones), and pestle and mortar are simple machines or employ principles of mechanical energy. For instance, the cutlass acts as a wedge and a lever. Modern agricultural machinery (tractors, tillers) also extensively uses principles of mechanical advantage, efficiency, and friction reduction. This allows for discussions on how science improves local practices.
Construction Industry: Cranes (complex machines utilizing pulleys and levers), wheelbarrows (lever, wheel and axle), and ramps (inclined planes) are ubiquitous on Nigerian construction sites. Knowledge of MA, VR, and efficiency helps engineers and builders select appropriate machinery, optimize work processes, and ensure safety.