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High School. PHYSICS. InClass. by SSL Technologies. with S. Lancione. Exercise-38. Kinetic Energy. Part 3 /3. EXERCISES. Question-1. A force of 12 N, acting 60 o from the horizontal, is applied to a 20 kg cart

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  1. High School PHYSICS InClass by SSL Technologies with S. Lancione Exercise-38 Kinetic Energy Part 3 /3

  2. EXERCISES

  3. Question-1 A force of 12 N, acting 60o from the horizontal, is applied to a 20 kg cart initially at rest resulting in a final velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 250 m). FA = 12 N vi = 0 vf = 10 m/s 60o f = 2 N 20 kg FH = 6 N B A s = 250 m No a) Was the cart at rest? “velocity not constant” (While traveling from point-A to point-B) 12 N [E 60o N] b) What was the applied force? 6 N right c) What was the horizontal component of the applied force? 2 N left d) What was the frictional force? 4 N right e) What was the resultant force? (FR = FA – f = 6 N – 2 N = 4 N) Click Click Click Click Click Click

  4. Question-1 A force of 12 N, acting 60o from the horizontal, is applied to a 20 kg cart initially at rest resulting in a final velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 250 m). FA = 12 N vi = 0 vf = 10 m/s 60o f = 2 N 20 kg FH = 6 N B A s = 250 m 0.2 m/s2 f) What was the acceleration of the cart? 0 g) What was the initial EK of the cart? 1000 J h) What was the final EK of the cart? 1000 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

  5. Question-1 Remember that the force must be parallel to the distance in the Work Formula. A force of 12 N, acting 60o from the horizontal, is applied to a 20 kg cart initially at rest resulting in a final velocity of 10 m/s. If the force of friction is 2 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 250 m). FA = 12 N vi = 0 vf = 10 m/s 60o f = 2 N 20 kg FH = 6 N B A s = 250 m 500 J k) How much work was done to overcome friction? 1500 J l) What was the total work done? n) Summarize the amounts of work done: 1000 J (1500 J – 500 J) 1) To accelerate the cart (Cart was not raised) 0 2) To raise the cart 500 J 3) To overcome friction 4) Total work done 1500 J Click Click Click Click Click Click Click Click Click Click

  6. Question-2 A force of 110 N, acting 60o from the horizontal, is applied to a 10 kg cart whose initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 24 m). FA = 110 N vi = 4 m/s vf = 16 m/s 60o f = 5 N 10 kg FH = 55 N B A s = 24 m No a) Was the cart at rest? “velocity not constant” (While traveling from point-A to point-B) 110 N [E 60o N] b) What was the applied force? 55 N right c) What was the horizontal component of the applied force? 5 N left d) What was the frictional force? 50 N right e) What was the resultant force? (FR = FA – f = 55 N – 5 N = 50 N) Click Click Click Click Click Click Click

  7. Question-2 A force of 110 N, acting 60o from the horizontal, is applied to a 10 kg cart whose initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 24 m). FA = 110 N vi = 4 m/s vf = 16 m/s 60o f = 5 N 10 kg FH = 55 N B A s = 24 m 5 m/s2 f) What was the acceleration of the cart? 80 J g) What was the initial EK of the cart? 1280 J h) What was the final EK of the cart? 1200 J i) How much work was done on the cart? Work done on cart = total energy – energy lost to friction 1280 J – 80 J = 1200 J Click Click Click Click Click

  8. Question-2 A force of 110 N, acting 60o from the horizontal, is applied to a 10 kg cart whose initial velocity is 4 m/s. The final velocity is 16 m/s. If the frictional force of 5 N, answer the following questions concerning the cart in going from point-A to point-B (a distance of 24 m). FA = 110 N vi = 4 m/s vf = 16 m/s 60o f = 5 N 10 kg FH = 55 N B A s = 24 m 120 J j) How much work was done to overcome friction? 1320 J k) What was the total work done? l) Summarize the amounts of work done: 1200 J (1320 J – 120 J) 1) To accelerate the cart (Cart was not raised) 0 2) To raise the cart 120 J 3) To overcome friction 4) Total work done 1320 J Click Click Click Click Click Click Click

  9. Question-3 A 10 kg cart is traveling at 12 m/s towards the right. If a frictional force of 40 N stops the cart in a distance of 18 m, answer the following questions concerning the cart in going from point-A to point-B. vi = 12 m/s vf = 0 f = 40 N 10 kg B A s = 18 m No a) Was the cart at rest? “velocity not constant” (While traveling from point-A to point-B) 40 N left (or – 40 N) b) What was the frictional force? 40 N left (or - 40 N) c) What was the resultant force? -4 m/s2 d) What was the acceleration? Click Click Click Click Click Click Click Click

  10. Question-3 A 10 kg cart is traveling at 12 m/s towards the right. If a frictional force of 40 N stops the cart in a distance of 18 m, answer the following questions concerning the cart in going from point-A to point-B. vi = 12 m/s vf = 0 f = 40 N 10 kg B A s = 18 m 720 J e) What was the initial EK of the cart? 0 f) What was the final EK of the cart? 720 J g) How much energy did the cart lose? h) What becomes of the energy lost by the cart? Used to overcome friction (lost as heat and sound) Click Click Click Click Click

  11. Question-4 A hammer falls from a scaffold and 1.5 s later strikes the ground with a kinetic energy of 157.5 J. What is the weight of the hammer? Answer Click

  12. Question-5 A projectile, whose mass is 800 g, is shot into the air with a velocity of 25 m/s, 42o N of E. Determine the kinetic energy of the projectile one secondafter it is fired. Answer Click Click

  13. Question-6 Starting from rest, a car reaches a velocity of 60 m/s in a distance of 120 m. Assuming the system is frictionless and knowing that the motor of the car produces a force of 3 x 104 N, calculate the mass of the car. Answer Click Click

  14. Question-7 The mass of an electron is 1.67 x 1027 kg. What work must be done on the electron in order to give it a speed of 2.5 x 107 m/s? Answer Click Click

  15. Question-8 A bullet of mass 2 g, traveling at 500 m/s, is fired at a piece of wood. The bullet emerges from the wood with a speed of 100 m/s. If the retarding force of friction was 4800 N, calculate the thickness of the piece of wood. ? Answer Click Click

  16. Question-9 What is the mass of a stone that is thrown in the air with avelocity of 1.1 m/s and with an initial kinetic energy of 0.0121 J? Answer Click Click

  17. Question-10 Two vehicles, X and Y, are traveling at the same speed. Vehicle-X has twice the kinetic energy of vehicle-Y. What is the value of the following ratio? Mass of vehicle-X Mass of vehicle-Y Answer Click Click

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