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High School by SSL Technologies Part 3 /3 Physics Ex-38 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 60o A FH = 6 N vf = 10 m/s f=2N s = 250 m 20 kg B No “velocity not constant” a) Was the cart at rest? (While traveling from point-A to point-B) 12 N [E 60o N] b) What was the applied force? c) What was the horizontal component of the applied force? 6 N right d) What was the frictional force? 2 N left e) What was the resultant force? 4 N right (FR = FA – f = 6 N – 2 N = 4 N) Click Physics Ex-38 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 60o A FH = 6 N vf = 10 m/s f=2N s = 250 m 20 kg B f) What was the acceleration of the cart? 0.2 m/s2 g) What was the initial EK of the cart? h) What was the final EK of the cart? 0 1 000 J i) How much work was done on the cart? 1000 J j) What becomes of the work done on the cart? It is transferred to the cart in the form of EK (faster speed). Click Physics Ex-38 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 60o A FH = 6 N f=2N s = 250 m vf = 10 m/s 20 kg B k) How much work was done to overcome friction? l) What was the total work done? 500 J 1500 J n) Summarize the amounts of work done: 1000 J (1500 J – 500 J) (Cart was not raised) 0 3) To overcome 500 be J parallel Remember that friction the force must 4)to Total done in the Work 1500Formula. J the work distance 1) To accelerate the cart 2) To raise the cart Click Physics Ex-38 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 60o A FH = 55 N vf = 16 m/s f=5N s = 24 m 10 kg B No “velocity not constant” a) Was the cart at rest? (While traveling from point-A to point-B) 110 N [E 60o N] b) What was the applied force? c) What was the horizontal component of the applied force? 55 N right d) What was the frictional force? 5 N left e) What was the resultant force? 50 N right (FR = FA – f = 55 N – 5 N = 50 N) Click Physics Ex-38 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 60o A FH = 55 N vf = 16 m/s f=5N s = 24 m 10 kg B f) What was the acceleration of the cart? g) What was the initial EK of the cart? 5 m/s2 80 J h) What was the final EK of the cart? 1 280 J i) How much work was done on the cart? 1 200 J Work done on cart = total energy – energy lost to friction 1280 J – 80 J = 1200 J Click Physics Ex-38 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 f=5N 60o A vf = 16 m/s FH = 55 N 10 kg s = 24 m B j) How much work was done to overcome friction? k) What was the total work done? 120 J 1320 J l) Summarize the amounts of work done: 1) To accelerate the cart 2) To raise the cart 3) To overcome friction 4) Total work done 1200 J 0 120 J 1320 J (1320 J – 120 J) (Cart was not raised) Click Physics Ex-38 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 A vf = 0 f = 40 N 10 kg s = 18 m B a) Was the cart at rest? No “velocity not constant” (While traveling from point-A to point-B) b) What was the frictional force? 40 N left (or – 40 N) c) What was the resultant force? 40 N left (or - 40 N) d) What was the acceleration? - 4 m/s2 Click Physics Ex-38 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 A vf = 0 f = 40 N 10 kg s = 18 m B e) What was the initial EK of the cart? f) What was the final EK of the cart? g) How much energy did the cart lose? 720 J 0 720 J h) What becomes of the energy lost by the cart? Used to overcome friction (lost as heat and sound) Click Click Question-4 Physics Ex-38 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? Click Question-5 Physics Ex-38 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 second after it is fired. Click Question-6 Physics Ex-38 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. Click Question-7 Physics Ex-38 The mass of an electron is 1.67 x 1027 kg. What work must be done on the electron in order to give it a speed of 2.5 x 107 m/s? Click Question-8 Physics Ex-38 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. ? Click Question-9 Physics Ex-38 What is the mass of a stone that is thrown in the air with a velocity of 1.1 m/s and with an initial kinetic energy of 0.0121 J? Click Question-10 Physics Ex-38 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 Click SSLTechnologies.com/science