Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com WORK-(KINETIC) ENERGY THEOREM With every click information will appear on the screen.

Similar presentations


Presentation on theme: "1 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com WORK-(KINETIC) ENERGY THEOREM With every click information will appear on the screen."— Presentation transcript:

1 1 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com WORK-(KINETIC) ENERGY THEOREM With every click information will appear on the screen. Use the information to answer questions on the worksheet. End of Slide

2 2 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com INDEX Work-Energy Energy Transferred Work Done Work – Newton 1 Work – Newton 2 Final Velocity Calculations Work-Energy Theorem Copyright Protection. © Direct-Science. All Rights Reserved worldwide under the Berne Convention. May not be copied or distributed without prior written permission from Direct-Science. If you have this file or a printout of it and it is not licensed in your name you are depriving the author and publisher of their rightful royalties. Please order your legal copy from the publisher at: http://chemistry-physics.com/ Thank you. Work-Energy Calculations End of Slide

3 3 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com  Work is done on an object when a force is acting on the object with the object being displaced in direction of  the applied force or  a component the applied force. Work Force Displacement  While Work is done on an object, energy is transferred to the object by the force acting on the object.  The energy change in the object allows the object to perform certain motions such as a,  positive acceleration,  negative acceleration,  or to maintain a constant velocity. Energy End of Slide

4 4 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com  The total amount or work done on an object or total amount energy transferred to the object by a force can be calculated with the following equation: W = F applied. ∆x. cos  [Work: Symbol – W ; Unit – Joule (J)]  [Force: Symbol – F ; Unit – Newton (N)]  [Change in Displacement: Symbol - ∆ x ; Unit – Meter (m)]  [Direction: Symbol - ; Direction of motion relative to direction of applied force]. Work Energy Force Direction Displacement End of Slide

5 5 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done  A force of 20 N is applied to an object with mass of 5 kg while the object is moving from rest a distance of 10 m over a frictionless surface in direction of the applied force. Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m  The amount of work done on the object can be calculated: W = F applied. ∆x. cos = 20 N. 10 m. cos 0 = 200 N.m.+1 = + 200 J (N.m) Direction Of Motion Direction Of Applied Force  Angle between direction of force and direction of motion = 0 0.  Amount of energy transferred to the object = 200 J.  The energy content of the object has increased by + 200 J. (+) (-) End of Slide

6 6 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m  The amount of energy transferred to the object will be converted by the object into other forms of energy such as  Heat in the presence of friction (to overcome friction)  Increase in Kinetic energy in which case the speed of the object will increase.  Decrease in Kinetic energy in which case the speed of the object will decrease. Direction Of Motion Direction Of Applied Force  The energy content of the object has increased by + 200 J. (+) (-) End of Slide

7 7 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m Direction Of Motion Direction Of Applied Force  The energy content of the object has increased by + 200 J.  The speed of the object will stay constant if the magnitude of applied force is equal to the magnitude of friction force: All energy transferred by the applied force to the object is used by object to overcome friction.  This observation is according to Newton's first Law of Motion: An object will stay at rest or move at a constant velocity unless acted upon by a non-zero resulted force. (+) (-) Newton First Law End of Slide

8 8 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m Direction Of Motion Direction Of Applied Force  The energy content of the object has increased by + 200 J.  The speed of the object will decrease if the magnitude of applied force is less than the magnitude of the friction force while the object is moving: All energy transferred by the applied force to the object as well as kinetic energy of the moving object is converted into heat energy and the speed of the object will decrease.  This observation is according to Newton's second Law of Motion: When a non-zero resulting force is acting on an object then the object will accelerate in direction of the force at an acceleration rate directly proportional to the magnitude of the resultant force and inversely proportional to the mass of the object. (+) (-) Newton Second Law End of Slide

9 9 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m Direction Of Motion Direction Of Applied Force  The energy content of the object has increased by + 200 J.  The speed of the object will increase if the magnitude of applied force is more than the magnitude of friction force while the object is moving: The resultant energy (excess energy after overcoming friction) will be converted into kinetic energy.  This observation is according to Newton's second Law of Motion: When a non-zero resulting force is acting on an object then the object will accelerate in direction of the force at an acceleration rate directly proportional to the magnitude of the resultant force and inversely proportional to the mass of the object. (+) (-) End of Slide

10 10 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m Direction Of Motion Direction Of Applied Force  The energy content of the object has increased by + 200 J.  The object is moving on a frictionless surface (does nor experience any friction).  All energy transferred by the object is converted into kinetic energy.  The speed of the object will increase.  The excess energy transferred to the object is obtained from the resultant force acting on the object.  The change in the speed of the object as a result of the energy transferred by resultant force can be calculated: W net = F net. ∆x. cos = ∆ E K = E k(f) – E k(i) = ½ mv 2 (f) – ½ mv 2 (i) (+) (-) End of Slide

11 11 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m Direction Of Motion Direction Of Applied Force  The energy content of the object has increased by + 200 J.  W net = F net. ∆x. cos = ∆ E K = E k(f) – E k(i) = ½ mv 2 (f) – ½ mv 2 (i) The speed of the object after 10 m of motion if it starts from rest (v i = 0 m.s -1 ) can be calculated: W net = F net. ∆x. cos = ∆ E K = E k(f) – E k(i) = ½ mv 2 (f) – ½ mv 2 (i) 200 J = ∆ E K = E k(f) – E k(i) = ½ mv 2 (f) – ½ mv 2 (i) 200 J = ½ mv 2 (f) – ½ mv 2 (i) 200 J = ½ mv 2 (f) – 0 J 200 J = ½. 5 kg. v 2 (f) 200 J = 2,5 kg. v 2 (f) v 2 (f) = 200 J / 2,5 kg v (f) = (200 J / 2,5 kg) 1/2 v (f) = 8,94 m.s -1 (+) (-) Velocity Calculation End of Slide

12 12 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com Work Calculate the Amount of Work Done Calculate the Amount of Energy Transferred 5 kg 20 N ∆x = 10 m Direction Of Motion Direction Of Applied Force  The final speed of v (f) = 8,94 m.s -1 of the moving object can be confirmed with equations of motion.  The acceleration of the moving object according to Newton's Second Law: a = F res / m = 20 N / 5 kg = 4 m.s -2. a = 4 m.s -2 ; v i = 0 m.s -1 ; ∆ x = 10 m. Equation of motion: v f 2 = v i 2 + 2. a. ∆x v f 2 = 0 + 2. 4. 10 = 80 v f = (80) 1/2 v f = 8,94 m.s -1 (+) (-) Final Speed Equation of Motion End of Slide

13 13 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com 13 End of Slide WORK-(KINETIC) ENERGY THEOREM : DOWNLOAD THE COMPLETE LECTURE FROM www.chemistry-physics.com HOME Download Lectures, Worksheets and Tutorials can be downloaded for any period of time.


Download ppt "1 PHYSICS TOPICS Index © DIRECT-SCIENCE : www.chemistry-physics.com WORK-(KINETIC) ENERGY THEOREM With every click information will appear on the screen."

Similar presentations


Ads by Google