Work Energy Theorem & Mechanical Energy Conservation.pptx
marcinejayar3
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Oct 12, 2025
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About This Presentation
Physics Lessons
Size: 112.71 KB
Language: en
Added: Oct 12, 2025
Slides: 15 pages
Slide Content
Work Energy Theorem & Mechanical Energy Conservation Presented By: Jay-ar B. Marcine STEM 12
Work Energy Theorem states that whenever work is done, energy changes. If work is done on an object, the net work is equal to its change in kinetic energy. This relationship can be written in the equation:
Example An 8.0-kilogram cart accelerates from 2.0 meters per second to 4.0 meters per second. What is the amount of work done? Given: M = 8.0 kg Vi = 2.0 m/s Vf = 4.0 m/s W = ?
Solution
Example A 10 kg box is pushed across a smooth floor. It accelerates from 1.5 m/s to 3.5 m/s. How much work was done on the box? Given: M = 10 kg Vi = 1.5 m/s Vf = 3.5 m/s W = ?
Mechanical Energy Conservation Mechanical energy is the total energy an object has due to: Kinetic Energy (KE) → motion Potential Energy (PE) → position or height Mechanical Energy (ME)=KE+PE
Conservation Principle: In a system without friction or air resistance , the total mechanical energy stays constant . That means energy can change forms (from PE to KE or vice versa), but the total amount doesn’t change. ME initial = ME initial or PE i = KE f + PE f
Real-Life Examples Falling Object (e.g., a mango from a tree 🍃) At the top: High PE, zero KE As it falls: PE decreases, KE increases Just before hitting the ground: All KE, no PE ✅ Total ME stays the same
Real-Life Examples Roller Coaster Ride 🎢 At the highest point: Maximum PE As it descends: PE converts to KE At the lowest point: Maximum KE ✅ Energy shifts, but total ME is conserved
Sample Problem A 2 kg ball is dropped from a height of 5 meters. What is its speed just before hitting the ground? Formula: PE – mgh ( mass, gravity and height ) KE – ½ mv ²( mass, velocity )
Solution Step1 : calculate the PE at the top PE = mgh = (2)(9.8)(5) = 98j Step 2: Set PE = KE at the bottom ½ mv ² =98 => ½ ( 2 ) v ²= 98 9.9 m/s
Sample Problem A 50 kg skater starts from rest at the top of a 3-meter-high ramp. Assuming no friction, what is the skater’s speed at the bottom of the ramp? Given: M = 50kg H = 3m