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InClass

High School. PHYSICS. InClass. by SSL Technologies. with S. Lancione. Exercise-37. Kinetic Energy. Part 2 /3. KINETIC ENERGY. PART-2. Energy is the ability to do work. There are three types of work that can be done:  Accelerating an object The work done goes to the object

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InClass

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  1. High School PHYSICS InClass by SSL Technologies with S. Lancione Exercise-37 Kinetic Energy Part 2 /3

  2. KINETIC ENERGY PART-2 Energy is the ability to do work. There are three types of work that can be done: Accelerating an objectThe work done goes to the object in the form of EK (kinetic energy) thereby increasing the velocity of the object. Raising an objectThe work done goes to the object in the form of EP (potential energy) thereby increasing the height of the object (from a reference line). To overcome frictionThe work done to overcome friction is “lost” as heat and sound energy. Click

  3. KINETIC ENERGY PART-2 An ideal system is one which has no friction. Frictionless systems do not exist in nature. However, since they simplify the solving of physics problems, it is often convenient to assume a system is frictionless. Once the problem is solved without friction, the effects caused by friction are added to the system. Remember that in an ideal system, there is no loss of energy due to friction. A real system is one that has friction. All systems in nature are real systems. That is, all systems in nature lose some energy due to friction (usually in the form of heat and sound). The Law of Conservation of Energy tells us that energy cannot be created nor destroyed. This means that when work is done, all of the energy must be accounted for. If the system is ideal, none of the work done is wasted. However, if the system is real, some of the work done is “lost” (as heat and sound energy). Click

  4. KINETIC ENERGY PART-2 When an object is accelerated, work is being done. When an object is decelerated, the kinetic energy it has must be dissipated (reduced to zero). When stopping a car, work is done by the brakes (friction) to dissipate the energy. The EK of the car is transformed into heat and sound. The fact that the kinetic energy of an object is proportional to the square of its speed has important consequences. Doubling the speed of a car, for example, quadruples its kinetic energy (22). If the velocity increases ten times, then the EK becomes 100 times greater (102). This explains the highway slogan that “speed kills”. Click

  5. EXERCISES

  6. Question-1 Explain the difference between an ideal system and a real system. An ideal system is one which has no friction and thus has no energy loss in the form of heat and sound. A real system is one which has friction and thus has energy loss in the form of heat and sound. Click Click

  7. Question-2 List the three types of work that can be done and tell what becomes of this work. Accelerating an object The work goes to the object in the form of EK. Raising an object The work goes to the object in the form of EP. To overcome friction The work is lost in the form of heat and sound energy. Click Click

  8. Question-3 A 2 kg cart rests motionless on a horizontal plane. Yes a) Is the cart at rest? 0 b) What is the resultant force? 0 c) What is the acceleration? 0 d) What is the kinetic energy? Click Click Click Click Click

  9. Question-4 A 2 kg cart travels at a constant velocity of 10 m/s from Point-A to Point-B, a distance of 5 m. Assuming the system is frictionless(f = 0), answer the following questions: vi = 10 m/s vf = 10 m/s 2 kg f = 0 B A s = 5 m Yes “constant velocity” a) Was the cart at rest? (While traveling from point-A to point-B) 0 b) What was the applied force? “given” 0 c) What was the frictional force? (FR = FA – f = 0 – 0 = 0) 0 d) What was the resultant force? “constant velocity” 0 e) What was the acceleration? Click Click Click Click Click Click

  10. Question-4 A 2 kg cart travels at a constant velocity of 10 m/s from Point-A to Point-B, a distance of 5 m. Assuming the system is frictionless(f = 0), answer the following questions: vi = 10 m/s vf = 10 m/s 2 kg f = 0 B A s = 5 m 100 J f) What was the initial EK? 100 J g) What was the final EK? 0 h) How much energy did the cart gain? 0 i) How much work was done on the cart? j) Summarize the amounts of work done: 0 1) To accelerate the cart 2) To raise the cart 0 0 3) To overcome friction 4) Total work done 0 Click Click Click Click Click Click Click Click Click Click Click

  11. Question-5 Starting from rest, a horizontal force of 20 N is applied to a 2 kg cart resulting in a final velocity of 10 m/s. Assuming the system is ideal (frictionless), answer the following questions. vi = 0 vf = 10 m/s FA = 20 N 2 kg f = 0 B A s = 5 m No a) Was the cart at rest? (While traveling from point-A to point-B) 20 N right b) What was the applied force? 20 N right c) What was the horizontal component of the applied force? “frictionless” 0 d) What was the frictional force? 20 N right e) What was the resultant force? (FR = FA – f = 20 N – 0 = 20 N) Click Click Click Click Click Click

  12. Question-5 Starting from rest, a horizontal force of 20 N is applied to a 2 kg cart resulting in a final velocity of 10 m/s. Assuming the system is ideal (frictionless), answer the following questions. vi = 0 vf = 10 m/s FA = 20 N 2 kg f = 0 B A s = 5 m 10 m/s2 f) What was the acceleration of the cart? 0 g) What was the initial EK of the cart? 100 J h) What was the final EK of the cart? 100 J i) How much work was done on the cart? j) What becomes of the work done on the cart? It is transferred to the cart in the form of EK (faster speed). Click Click Click Click Click Click Click Click Click Click

  13. Question-5 Starting from rest, a horizontal force of 20 N is applied to a 2 kg cart resulting in a final velocity of 10 m/s. Assuming the system is ideal (frictionless), answer the following questions. vi = 0 vf = 10 m/s FA = 20 N 2 kg f = 0 B A s = 5 m 0 k) How much work was done to overcome friction? 100 J l) What was the totalwork done? The applied force m) Where did the work done come from? n) Summarize the amounts of work done: 100 J 1) To accelerate the cart (Cart was not raised) 2) To raise the cart 0 0 3) To overcome friction (Frictionless system) 4) Total work done 100 J Click Click Click Click Click Click Click Click Click Click Click

  14. Question-6 A horizontal force of 20 N is applied to a 2 kg cart whose initial velocity is 8 m/s resulting in a final velocity of 12 m/s. Assuming there is no friction, answer the following questions concerning the cart in going from Point-A to Point-B (4 m). vi = 8 m/s vf = 12 m/s FA = 20 N 2 kg f = 0 B A s = 4 m Was the cart at rest? No a) (While traveling from point-A to point-B) 20 N right b) What was the applied force? 20 N right c) What was the horizontal component of the applied force? “frictionless” 0 d) What was the frictional force? 20 N right e) What was the resultant force? (FR = FA – f = 20 N – 0 = 20 N) Click Click Click Click Click Click

  15. Question-6 A horizontal force of 20 N is applied to a 2 kg cart whose initial velocity is 8 m/s resulting in a final velocity of 12 m/s. Assuming there is no friction, answer the following questions concerning the cart in going from Point-A to Point-B (4 m). vi = 8 m/s vf = 12 m/s FA = 20 N 2 kg f = 0 B A s = 4 m 10 m/s2 f) What was the acceleration of the cart? 64 J g) What was the initial EK of the cart? 144 J h) What was the final EK of the cart? 80 J i) How much work was done on the cart? j) What becomes of the work done on the cart? It is transferred to the cart in the form of EK (faster speed). Click Click Click Click Click Click Click Click Click Click

  16. Question-6 A horizontal force of 20 N is applied to a 2 kg cart whose initial velocity is 8 m/s resulting in a final velocity of 12 m/s. Assuming there is no friction, answer the following questions concerning the cart in going from Point-A to Point-B (4 m). vi = 8 m/s vf = 12 m/s FA = 20 N 2 kg f = 0 B A s = 4 m 0 k) How much work was done to overcome friction? 80 J l) What was the totalwork done? The applied force m) Where did the work done come from? n) Summarize the amounts of work done: 80 J 1) To accelerate the cart (Cart was not raised) 2) To raise the cart 0 0 3) To overcome friction (Frictionless system) 4) Total work done 80 J Click Click Click Click Click Click Click Click Click Click

  17. Question-7 A 2 kg cart travels at a constant velocity of 10 m/s for a distance of 5 m. If the frictional force is 2 N, answer the following questions concerning the cart in going from Point-A to Point-B. vi = 10 m/s vf = 10 m/s FA = ? 2 kg f = 2 N B A s = 5 m “constant velocity” Was the cart at rest? Yes a) (While traveling from point-A to point-B) 2 N right b) What was the applied force? 2 N right c) What was the horizontal component of the applied force? 2 N left d) What was the frictional force? 0 e) What was the resultant force? (FR = FA – f = 2 N – 2 N = 0) Click Click Click Click Click Click Click Click

  18. Question-7 A 2 kg cart travels at a constant velocity of 10 m/s for a distance of 5 m. If the frictional force is 2 N, answer the following questions concerning the cart in going from Point-A to Point-B. vi = 10 m/s vf = 10 m/s FA = ? 2 kg f = 2 N B A s = 5 m 0 f) “constant velocity” What was the acceleration of the cart? 100 J g) What was the initial EK of the cart? 100 J h) What was the final EK of the cart? 0 i) How much work was done on the cart? j) What becomes of the work done? It is used to overcome friction (lost energy). Click Click Click Click Click Click Click Click Click Click

  19. Question-7 A 2 kg cart travels at a constant velocity of 10 m/s for a distance of 5 m. If the frictional force is 2 N, answer the following questions concerning the cart in going from Point-A to Point-B. A 2 kg cart travels at a constant velocity of 10 m/s for a distance of 5 m. If the frictional force is 2 N, answer the following questions concerning the cart in going from Point-A to Point-B. vi = 10 m/s vf = 10 m/s FA = ? 2 kg f = 2 N B A s = 5 m 10 J k) How much work was done to overcome friction? 10 J l) What was the totalwork done? The applied force m) Where did the work done come from? n) Summarize the amounts of work done: (Cart was not accelerated) 0 1) To accelerate the cart (Cart was not raised) 2) To raise the cart 0 10 J 3) To overcome friction 4) Total work done 10 J Click Click Click Click Click Click Click Click Click Click

  20. Question-8 A horizontal force of 20 N is applied to a 2 kg cart, initially at rest, giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer the following questions concerning the cart in going from A to B. vi = 0 vf = 27 m/s FA = 20 N 2 kg f = 2 N B A s = 40.5 m Was the cart at rest? No “velocity not constant” a) (While traveling from point-A to point-B) 20 N right b) What was the applied force? 20 N right c) What was the horizontal component of the applied force? 2 N left d) What was the frictional force? 18 N right e) What was the resultant force? (FR = FA – f = 20 N – 2 N = 18 N) Click Click Click Click Click Click Click Click

  21. Question-8 A horizontal force of 20 N is applied to a 2 kg cart, initially at rest, giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer the following questions concerning the cart in going from A to B. vi = 0 vf = 27 m/s FA = 20 N 2 kg f = 2 N B A s = 40.5 m 9 m/s2 f) What was the acceleration of the cart? 0 g) What was the initial EK of the cart? 729 J h) What was the final EK of the cart? 729 J i) How much work was done on the cart? j) What becomes of the work done on the cart? It is used to accelerate the cart(increase its velocity from 0 to 27 m/s). Click Click Click Click Click Click Click Click Click Click

  22. Question-8 A horizontal force of 20 N is applied to a 2 kg cart, initially at rest, giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer the following questions concerning the cart in going from A to B. vi = 0 vf = 27 m/s FA = 20 N 2 kg f = 2 N B A s = 40.5 m 81 J k) How much work was done to overcome friction? 810 J l) What was the totalwork done? The applied force m) Where did the work done come from? n) Summarize the amounts of work done: 729 J 1) To accelerate the cart (Cart was not raised) 2) To raise the cart 0 81 J 3) To overcome friction 4) Total work done 810 J Click Click Click Click Click Click Click Click Click Click

  23. Question-9 A horizontal force of 20 N is applied to a 2 kg cart, initially at rest, giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer the following questions concerning the cart in going from A to B. vi = 6 m/s vf = 18 m/s FA = 20 N 2 kg f = 2 N B A s = 16 m Was the cart at rest? No “velocity not constant” a) (While traveling from point-A to point-B) 20 N right b) What was the applied force? 20 N right c) What was the horizontal component of the applied force? 2 N left d) What was the frictional force? 18 N right e) What was the resultant force? (FR = FA – f = 20 N – 2 N = 18 N) Click Click Click Click Click Click Click Click

  24. Question-9 A horizontal force of 20 N is applied to a 2 kg cart, initially at rest, giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer the following questions concerning the cart in going from A to B. vi = 6 m/s vf = 18 m/s FA = 20 N 2 kg f = 2 N B A s = 16 m 9 m/s2 f) What was the acceleration of the cart? 36 J g) What was the initial EK of the cart? 324 J h) What was the final EK of the cart? 288 J i) How much work was done on the cart? j) What becomes of the work done on the cart? It is used to accelerate the cart(increase its velocity from 6 to 18 m/s). Click Click Click Click Click Click Click Click Click Click

  25. Question-9 A horizontal force of 20 N is applied to a 2 kg cart, initially at rest, giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer the following questions concerning the cart in going from A to B. vi = 6 m/s vf = 18 m/s FA = 20 N 2 kg f = 2 N B A s = 16 m 32 J k) How much work was done to overcome friction? 320 J l) What was the totalwork done? The applied force m) Where did the work done come from? n) Summarize the amounts of work done: 288 J 1) To accelerate the cart (Cart was not raised) 2) To raise the cart 0 32 J 3) To overcome friction 4) Total work done 320 J Click Click Click Click Click Click Click Click Click Click

  26. Question-10 A force is applied, at 60o from the horizontal, to a 20 kg cart causing it to travel at a constant velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from A to B (5 m). FA = ? vi = 10 m/s vf = 10 m/s 60o f = 2 N 20 kg FH = 2 N B A s = 5 m Was the cart at rest? Yes “velocity is constant” a) (While traveling from point-A to point-B) 4 N [E 60o N] b) What was the applied force? 2 N right c) What was the horizontal component of the applied force? 2 N left d) What was the frictional force? 0 e) What was the resultant force? (FR = FA – f = 2 N – 2 N = 0) Click Click Click Click Click Click Click Click Click

  27. Question-10 A force is applied, at 60o from the horizontal, to a 20 kg cart causing it to travel at a constant velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from A to B (5 m). FA = ? vi = 10 m/s vf = 10 m/s 60o f = 2 N 20 kg FH = 2 N B A s = 5 m 0 f) “velocity constant” What was the acceleration of the cart? 1000 J g) What was the initial EK of the cart? 1000 J h) What was the final EK of the cart? 0 i) How much work was done on the cart? j) What becomes of the work done on the cart? It is used to overcome friction. Click Click Click Click Click Click Click Click Click Click

  28. Question-10 A force is applied, at 60o from the horizontal, to a 20 kg cart causing it to travel at a constant velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from A to B (5 m). FA = ? vi = 10 m/s vf = 10 m/s 60o f = 2 N 20 kg FH = 2 N B A s = 5 m 10 J k) How much work was done to overcome friction? 10 J l) What was the totalwork done? The applied force m) Where did the work done come from? n) Summarize the amounts of work done: (Cart was not accelerated) 0 1) To accelerate the cart (Cart was not raised) 2) To raise the cart 0 10 J 3) To overcome friction 4) Total work done 10 J Click Click Click Click Click Click Click Click Click Click

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