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Part 2 /3 High School by SSL Technologies Physics Ex-37 Click Energy is the ability to do work. There are three types of work that can be done:  Accelerating.

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Presentation on theme: "Part 2 /3 High School by SSL Technologies Physics Ex-37 Click Energy is the ability to do work. There are three types of work that can be done:  Accelerating."— Presentation transcript:

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2 Part 2 /3 High School by SSL Technologies

3 Physics Ex-37 Click 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 in the form of E K (kinetic energy) thereby increasing the velocity of the object.  Raising an object The work done goes to the object in the form of E P (potential energy) thereby increasing the height of the object (from a reference line).  To overcome friction The work done to overcome friction is “lost” as heat and sound energy. PART-2

4 Physics Ex-37 Click 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). PART-2

5 Physics Ex-37 Click 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 E K 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 (2 2 ). If the velocity increases ten times, then the E K becomes 100 times greater (10 2 ). This explains the highway slogan that “speed kills”. PART-2

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7 Physics Ex-37 Question-1 Explain the difference between an ideal system and a real system. Click 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.

8 Physics Ex-37 Question-2 List the three types of work that can be done and tell what becomes of this work. Click Accelerating an object The work goes to the object in the form of E K. Raising an object The work goes to the object in the form of E P. To overcome friction The work is lost in the form of heat and sound energy.

9 Physics Ex-37 Question-3 A 2 kg cart rests motionless on a horizontal plane. Click Is the cart at rest? a) Yes What is the resultant force? b) 0 What is the acceleration? c) 0 What is the kinetic energy? d) 0 Click

10 Physics Ex-37 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: Click Was the cart at rest? a) Yes What was the applied force? b) 0 What was the frictional force? c) 0 What was the resultant force? d) 0 Click s = 5 m “given” v i = 10 m/sv f = 10 m/s What was the acceleration? e) 0 2 kg Click f = 0 “constant velocity” A B (While traveling from point-A to point-B) (F R = F A – f = 0 – 0 = 0) “constant velocity”

11 Physics Ex-37 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: Click What was the initial E K ? f) 100 J What was the final E K ? g) 100 J How much energy did the cart gain? h) 0 How much work was done on the cart? i) 0 Click s = 5 m v i = 10 m/sv f = 10 m/s Click Summarize the amounts of work done: j) 2 kg Click f = 0 A B 1) To accelerate the cart 0 2) To raise the cart 0 3) To overcome friction 0 4) Total work done 0 Click

12 Physics Ex-37 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. Click Was the cart at rest? a) No What was the applied force? b) 20 N right What was the horizontal component of the applied force? c) 20 N right What was the frictional force? d) 0 Click s = 5 m “frictionless” v i = 0v f = 10 m/s What was the resultant force? e) 20 N right 2 kg Click f = 0 A B (While traveling from point-A to point-B) (F R = F A – f = 20 N – 0 = 20 N) F A = 20 N

13 Physics Ex-37 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. Click What was the acceleration of the cart? f) 10 m/s 2 What was the initial E K of the cart? g) 0 What was the final E K of the cart? h) 100 J How much work was done on the cart? i) 100 J Click s = 5 m v i = 0v f = 10 m/s What becomes of the work done on the cart? j) It is transferred to the cart in the form of E K (faster speed). 2 kg Click f = 0 A B F A = 20 N Click

14 Physics Ex-37 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. Click How much work was done to overcome friction? k) 0 What was the total work done? l) 100 J Where did the work done come from? m) The applied force Summarize the amounts of work done: n) Click s = 5 m v i = 0v f = 10 m/s 2 kg f = 0 A B F A = 20 N Click 1) To accelerate the cart 100 J 2) To raise the cart 0 3) To overcome friction 0 4) Total work done 100 J Click (Cart was not raised) (Frictionless system) Click

15 Physics Ex-37 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). Click Was the cart at rest? a) No What was the applied force? b) 20 N right What was the horizontal component of the applied force? c) 20 N right What was the frictional force? d) 0 Click s = 4 m “frictionless” v i = 8 m/sv f = 12 m/s What was the resultant force? e) 20 N right 2 kg Click f = 0 A B (While traveling from point-A to point-B) (F R = F A – f = 20 N – 0 = 20 N) F A = 20 N

16 Physics Ex-37 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). s = 4 m v i = 8 m/sv f = 12 m/s 2 kg f = 0 A B F A = 20 N What was the acceleration of the cart? f) 10 m/s 2 What was the initial E K of the cart? g) 64 J What was the final E K of the cart? h) 144 J How much work was done on the cart? i) 80 J What becomes of the work done on the cart? j) It is transferred to the cart in the form of E K (faster speed). Click

17 Physics Ex-37 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). s = 4 m v i = 8 m/sv f = 12 m/s 2 kg f = 0 A B F A = 20 N How much work was done to overcome friction? k) 0 What was the total work done? l) 80 J Where did the work done come from? m) The applied force Summarize the amounts of work done: n) 1) To accelerate the cart 80 J 2) To raise the cart 0 3) To overcome friction 0 4) Total work done 80 J (Cart was not raised) (Frictionless system) Click

18 Physics Ex-37 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. Click Was the cart at rest? a) Yes What was the applied force? b) 2 N right What was the horizontal component of the applied force? c) 2 N right What was the frictional force? d) 2 N left Click s = 5 m v i = 10 m/sv f = 10 m/s What was the resultant force? e) 0 2 kg Click f = 2 N A B (While traveling from point-A to point-B) (F R = F A – f = 2 N – 2 N = 0) F A = ? Click “constant velocity”

19 Physics Ex-37 Question-7 What was the acceleration of the cart? f) 0 What was the initial E K of the cart? g) 100 J What was the final E K of the cart? h) 100 J How much work was done on the cart? i) 0 What becomes of the work done? j) It is used to overcome friction (lost energy). 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. s = 5 m v i = 10 m/sv f = 10 m/s 2 kg f = 2 N A B F A = ? “constant velocity” Click

20 Physics Ex-37 Question-7 How much work was done to overcome friction? k) 10 J What was the total work done? l) 10 J Where did the work done come from? m) The applied force Summarize the amounts of work done: n) 1) To accelerate the cart 0 2) To raise the cart 0 3) To overcome friction 10 J 4) Total work done 10 J (Cart was not raised) Click 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. s = 5 m v i = 10 m/sv f = 10 m/s 2 kg f = 2 N A B F A = ? (Cart was not accelerated)

21 Physics Ex-37 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. Click Was the cart at rest? a) No What was the applied force? b) 20 N right What was the horizontal component of the applied force? c) 20 N right What was the frictional force? d) 2 N left Click s = 40.5 m v i = 0v f = 27 m/s What was the resultant force? e) 18 N right 2 kg Click f = 2 N A B (While traveling from point-A to point-B) (F R = F A – f = 20 N – 2 N = 18 N) F A = 20 N Click “velocity not constant”

22 Physics Ex-37 What was the acceleration of the cart? f) 9 m/s 2 What was the initial E K of the cart? g) 0 What was the final E K of the cart? h) 729 J How much work was done on the cart? i) 729 J What becomes of the work done on the cart? j) It is used to accelerate the cart (increase its velocity from 0 to 27 m/s). 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. s = 40.5 m v i = 0v f = 27 m/s 2 kg f = 2 N A B F A = 20 N Click

23 Physics Ex-37 Question-8 How much work was done to overcome friction? k) 81 J What was the total work done? l) 810 J Where did the work done come from? m) The applied force Summarize the amounts of work done: n) 1) To accelerate the cart 729 J 2) To raise the cart 0 3) To overcome friction 81 J 4) Total work done 810 J (Cart was not raised) Click 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. s = 40.5 m v i = 0v f = 27 m/s 2 kg f = 2 N A B F A = 20 N

24 Physics Ex-37 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. Click Was the cart at rest? a) No What was the applied force? b) 20 N right What was the horizontal component of the applied force? c) 20 N right What was the frictional force? d) 2 N left Click s = 16 m v i = 6 m/sv f = 18 m/s What was the resultant force? e) 18 N right 2 kg Click f = 2 N A B (While traveling from point-A to point-B) (F R = F A – f = 20 N – 2 N = 18 N) F A = 20 N Click “velocity not constant”

25 Physics Ex-37 What was the acceleration of the cart? f) 9 m/s 2 What was the initial E K of the cart? g) 36 J What was the final E K of the cart? h) 324 J How much work was done on the cart? i) 288 J What becomes of the work done on the cart? j) It is used to accelerate the cart (increase its velocity from 6 to 18 m/s). Click 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. s = 16 m v i = 6 m/sv f = 18 m/s 2 kg f = 2 N A B F A = 20 N

26 Physics Ex-37 How much work was done to overcome friction? k) 32 J What was the total work done? l) 320 J Where did the work done come from? m) The applied force Summarize the amounts of work done: n) 1) To accelerate the cart 288 J 2) To raise the cart 0 3) To overcome friction 32 J 4) Total work done 320 J (Cart was not raised) Click 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. s = 16 m v i = 6 m/sv f = 18 m/s 2 kg f = 2 N A B F A = 20 N

27 Physics Ex-37 Question-10 A force is applied, at 60 o 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). Click Was the cart at rest? a) Yes What was the applied force? b) 4 N [E 60 o N] What was the horizontal component of the applied force? c) 2 N right What was the frictional force? d) 2 N left Click s = 5 m v i = 10 m/sv f = 10 m/s What was the resultant force? e) 0 20 kg Click f = 2 N A B (While traveling from point-A to point-B) (F R = F A – f = 2 N – 2 N = 0) F A = ? Click “velocity is constant” 60 o F H = 2 N Click

28 Physics Ex-37 What was the acceleration of the cart? f) 0 What was the initial E K of the cart? g) J What was the final E K of the cart? h) J How much work was done on the cart? i) 0 What becomes of the work done on the cart? j) It is used to overcome friction. Click Question-10 A force is applied, at 60 o 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). s = 5 m v i = 10 m/sv f = 10 m/s 20 kg f = 2 N A B F A = ? 60 o F H = 2 N “velocity constant”

29 Physics Ex-37 How much work was done to overcome friction? k) 10 J What was the total work done? l) 10 J Where did the work done come from? m) The applied force Summarize the amounts of work done: n) 1) To accelerate the cart 0 2) To raise the cart 0 3) To overcome friction 10 J 4) Total work done 10 J (Cart was not raised) Click Question-10 A force is applied, at 60 o 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). s = 5 m v i = 10 m/sv f = 10 m/s 20 kg f = 2 N A B F A = ? 60 o F H = 2 N (Cart was not accelerated)

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