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App of the 2nd law single body problems

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Nestor Enriquez
Nestor Enriquez

1. Identify and define the system of interest and draw a labeled diagram showing all relevant bodies and forces. 2. Write out Newton's second law equation for each body. 3. Decompose all forces acting on each body into their x- and y-components. 4. Sum the forces in the x- and y-directions for each body. 5. Write the system of equations and solve simultaneously to find the unknown accelerations. 6. Check units and answer the question asked.

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Lets recall:How is the second law of motion represented mathematically?What does the unit newton mean?
Application of the Second LawSingle-Body Problems
In which direction is the net force acting car A when it is moving east? A In which direction is the net force acting car B when it is braking to a stop while moving east? B
PROBLEM-SOLVINGTECHNIQUES
Mixed Up Recipe
Indicate all given information and state what is to be found.
Substitute all given quantities and solve for the unknown.
From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.
Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = maRead the problem carefully and then draw and label a rough sketch.Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.
Indicate a consistent positive direction along the continuous line of motion.Read the problem carefully and then draw and label a rough sketch.2. Indicate all given information and state what is to be found.3. Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.4. Indicate a consistent positive direction along the continuous line of motion.5. From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.6. Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = ma7. Substitute all given quantities and solve for the unknown.
Illustrative Example no.11. A 1000-kg car travels on a straight highway with a speed of 30.0 m/s. The driver sees a red light ahead and applies her brakes which exerts a constant braking force of 4.0kN. What is the acceleration of the car?m = 1000 kg1a = ?v = 30.0 m/s2F = 4.0kN = 4000N
Draw a Free-Body Diagram3yF = 4000 Nxa
Rightward Positive4y5Fnet = - 4000 NFnet = ma6F = - 4000 NFnetmxa = -4000 N1000 kga = ?7a = a = -4.0 m/s2
Illustrative Example no.22. A 15.0-kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. Neglecting friction, what is the magnitude and direction of the acceleration of the box? m F11F2 a 2 a = ? Given: Find: F1 =15.0 NF2 =18.0 N
Draw a Free-Body Diagram3y2. A 15.0-kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box?Fnet = F1 + F2xam = 15.0 kg
Rightward positive45Fnet = F1 + F2yFnet = 15.0N + 18.0N = 33.0NFnet = ma6Fnet= +Fnetma = xa = +m = 15.0 kg 33.0 N15.0 kg7a = a = ?a = 2.20 m/s2, Right a = +2.20 m/s2
Illustrative Example no.33. A freight elevator is lifted upward with an acceleration of 2.5 m/s2. If the tension in the supporting cable is 9600 N, what is the mass of the elevator and its contents?1TamGiven: a = 2.5 m/s2; T = 9600 N2Find: m = ?W
Draw a Free-Body Diagram3y3. A freight elevator is lifted upward with an acceleration of 2.5 m/s2. If the tension in the supporting cable is 9600 N, what is the mass of the elevator and its contents?T axW Weight = mass x acceleration due to gravityW = mg
Upward, positive45Fnet = T mgT mg = may6T = ma + mg= m(a + g)T(a + g)T = +m = a = +9600 N2.5 m/s2 + 9.8 m/s2 m = 7xm = ?9600 N12.3 m/s2 m = W = - mg m = 780 kg
Solve the followingWhat horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?
What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?a1Pmf2Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?
What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?+a3f-Free-body diagramP+4Rightward +5Fnet = P f
What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?5Fnet = P f P = 24.0 N + 20.0 NP f = ma 6Equate Fnet to ma, Fnet = maP = 44.0 N P = ma + f Derive equation to find the unknown7P = (6.0 kg)( 4.0 m/s2)+ 20.0 N
2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?1Tam2Given: a = 5.0 m/s2; ` m = 100 kgWFind: T = ?
2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?3T+a+W-4Upward (+)5Fnet = T W = T mg
2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?T+a+5Fnet = T mg WT mg = ma-6T = mg + maT = m(g + a)7T = (100 kg)(9.8 m/s2+ 5.0 m/s2) T = 100kg(14.8 m/s2)= 1480 kg
Solve in paper number 4.It is determined that the net force of 60.0 N will give a cart an acceleration of 10 m/s2. What force is required to give the same cart an acceleration of (a) 2.0 m/s2? (b) 5.0 m/s2 and (c) 15.0 m/s2 ? Ans. (a)F = 12.0 N; (b) 30,0 N and F =90.0 N 2. A 20.0-kg mass hangs at the end of a rope. Find the acceleration of the mass if the tension in the rope is (a) 196 N; (b) 120 N; and (c) 260 N. Ans. (a) 0; (b) -3.8 m/s2; (c) + 3.2 m/s2
3. A 10-kg mass is lifted upward by light cable. What is the tension in the cable if the acceleration is (a) zero, (b) 6.0 m/s2 upward, and (c) 6.0 m/s2 downward?4. An 800-kg elevator is lifted vertically by strong rope. Find the acceleration of the elevator if the rope tension is (a) 9000 N, (b) 7840 N, and (c) 2000 N?

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App of the 2nd law single body problems

  • 1. Lets recall:How is the second law of motion represented mathematically?What does the unit newton mean?
  • 2. Application of the Second LawSingle-Body Problems
  • 3. In which direction is the net force acting car A when it is moving east? A In which direction is the net force acting car B when it is braking to a stop while moving east? B
  • 6. Indicate all given information and state what is to be found.
  • 7. Substitute all given quantities and solve for the unknown.
  • 8. From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.
  • 9. Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = maRead the problem carefully and then draw and label a rough sketch.Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.
  • 10. Indicate a consistent positive direction along the continuous line of motion.Read the problem carefully and then draw and label a rough sketch.2. Indicate all given information and state what is to be found.3. Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.4. Indicate a consistent positive direction along the continuous line of motion.5. From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.6. Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = ma7. Substitute all given quantities and solve for the unknown.
  • 11. Illustrative Example no.11. A 1000-kg car travels on a straight highway with a speed of 30.0 m/s. The driver sees a red light ahead and applies her brakes which exerts a constant braking force of 4.0kN. What is the acceleration of the car?m = 1000 kg1a = ?v = 30.0 m/s2F = 4.0kN = 4000N
  • 12. Draw a Free-Body Diagram3yF = 4000 Nxa
  • 13. Rightward Positive4y5Fnet = - 4000 NFnet = ma6F = - 4000 NFnetmxa = -4000 N1000 kga = ?7a = a = -4.0 m/s2
  • 14. Illustrative Example no.22. A 15.0-kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. Neglecting friction, what is the magnitude and direction of the acceleration of the box? m F11F2 a 2 a = ? Given: Find: F1 =15.0 NF2 =18.0 N
  • 15. Draw a Free-Body Diagram3y2. A 15.0-kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box?Fnet = F1 + F2xam = 15.0 kg
  • 16. Rightward positive45Fnet = F1 + F2yFnet = 15.0N + 18.0N = 33.0NFnet = ma6Fnet= +Fnetma = xa = +m = 15.0 kg 33.0 N15.0 kg7a = a = ?a = 2.20 m/s2, Right a = +2.20 m/s2
  • 17. Illustrative Example no.33. A freight elevator is lifted upward with an acceleration of 2.5 m/s2. If the tension in the supporting cable is 9600 N, what is the mass of the elevator and its contents?1TamGiven: a = 2.5 m/s2; T = 9600 N2Find: m = ?W
  • 18. Draw a Free-Body Diagram3y3. A freight elevator is lifted upward with an acceleration of 2.5 m/s2. If the tension in the supporting cable is 9600 N, what is the mass of the elevator and its contents?T axW Weight = mass x acceleration due to gravityW = mg
  • 19. Upward, positive45Fnet = T mgT mg = may6T = ma + mg= m(a + g)T(a + g)T = +m = a = +9600 N2.5 m/s2 + 9.8 m/s2 m = 7xm = ?9600 N12.3 m/s2 m = W = - mg m = 780 kg
  • 20. Solve the followingWhat horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?
  • 21. What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?a1Pmf2Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?
  • 22. What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?+a3f-Free-body diagramP+4Rightward +5Fnet = P f
  • 23. What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?5Fnet = P f P = 24.0 N + 20.0 NP f = ma 6Equate Fnet to ma, Fnet = maP = 44.0 N P = ma + f Derive equation to find the unknown7P = (6.0 kg)( 4.0 m/s2)+ 20.0 N
  • 24. 2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?1Tam2Given: a = 5.0 m/s2; ` m = 100 kgWFind: T = ?
  • 25. 2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?3T+a+W-4Upward (+)5Fnet = T W = T mg
  • 26. 2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?T+a+5Fnet = T mg WT mg = ma-6T = mg + maT = m(g + a)7T = (100 kg)(9.8 m/s2+ 5.0 m/s2) T = 100kg(14.8 m/s2)= 1480 kg
  • 27. Solve in paper number 4.It is determined that the net force of 60.0 N will give a cart an acceleration of 10 m/s2. What force is required to give the same cart an acceleration of (a) 2.0 m/s2? (b) 5.0 m/s2 and (c) 15.0 m/s2 ? Ans. (a)F = 12.0 N; (b) 30,0 N and F =90.0 N 2. A 20.0-kg mass hangs at the end of a rope. Find the acceleration of the mass if the tension in the rope is (a) 196 N; (b) 120 N; and (c) 260 N. Ans. (a) 0; (b) -3.8 m/s2; (c) + 3.2 m/s2
  • 28. 3. A 10-kg mass is lifted upward by light cable. What is the tension in the cable if the acceleration is (a) zero, (b) 6.0 m/s2 upward, and (c) 6.0 m/s2 downward?4. An 800-kg elevator is lifted vertically by strong rope. Find the acceleration of the elevator if the rope tension is (a) 9000 N, (b) 7840 N, and (c) 2000 N?
  • 29. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?yFNxamWx30属W
  • 30. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?yFNxFnet = Wx = W sin 30属 Fnet= mg sin 30 属amg sin 30 属 = maWx30属g sin 30 属= a9.8 m/s2sin 30 属= aa = 4.9 m/s2W
  • 31. AssignmentDevise your own problem-solving strategy in solving problems in multiple-body systems in the application of the second law of motion?