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App of the 2nd law single body problems

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  • 1. Let’s recall:
    How is the second law of motion represented mathematically?
    What does the unit “newton” mean?
  • 2. Application of the Second Law
    Single-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
  • 4. PROBLEM-SOLVING
    TECHNIQUES
  • 5. Mixed Up Recipe
  • 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 = ma
    Read 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 = ma
    7. Substitute all given quantities and solve for the unknown.
  • 11. Illustrative Example no.1
    1. 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 kg
    1
    a = ?
    v = 30.0 m/s
    2
    F = 4.0kN = 4000N
  • 12. Draw a Free-Body Diagram
    3
    y
    F = 4000 N
    x
    a
  • 13. Rightward Positive
    4
    y
    5
    Fnet = - 4000 N
    Fnet = ma
    6
    F = - 4000 N
    Fnet
    m
    x
    a =
    -4000 N
    1000 kg
    a = ?
    7
    a =
    a = -4.0 m/s2
  • 14. Illustrative Example no.2
    2. 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
    F1
    1
    F2
    a
    2
    a = ?
    Given:
    Find:
    F1 =15.0 N
    F2 =18.0 N
  • 15. Draw a Free-Body Diagram
    3
    y
    2. 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 + F2
    x
    a
    m = 15.0 kg
  • 16. Rightward positive
    4
    5
    Fnet = F1 + F2
    y
    Fnet = 15.0N + 18.0N = 33.0N
    Fnet = ma
    6
    Fnet
    = +
    Fnet
    m
    a =
    x
    a
    = +
    m = 15.0 kg
    33.0 N
    15.0 kg
    7
    a =
    a = ?
    a = 2.20 m/s2, Right
    a = +2.20 m/s2
  • 17. Illustrative Example no.3
    3. 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?
    1
    T
    a
    m
    Given: a = 2.5 m/s2; T = 9600 N
    2
    Find: m = ?
    W
  • 18. Draw a Free-Body Diagram
    3
    y
    3. 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
    a
    x
    W
    Weight = mass x acceleration due to gravity
    W = mg
  • 19. Upward, positive
    4
    5
    Fnet = T – mg
    T – mg = ma
    y
    6
    T = ma + mg
    = m(a + g)
    T
    (a + g)
    T
    = +
    m =
    a
    = +
    9600 N
    2.5 m/s2 + 9.8 m/s2
    m =
    7
    x
    m = ?
    9600 N
    12.3 m/s2
    m =
    W = - mg
    m = 780 kg
  • 20. Solve the following
    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?
    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?
    a
    1
    P
    m
    f
    2
    Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 N
    Find: 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 N
    Find: P = ?
    +
    a
    3
    f
    -
    Free-body diagram
    P
    +
    4
    Rightward +
    5
    Fnet = 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 N
    Find: P = ?
    5
    Fnet = P – f
    P = 24.0 N + 20.0 N
    P – f = ma
    6
    Equate Fnet to ma, Fnet = ma
    P = 44.0 N
    P = ma + f
    Derive equation to find the unknown
    7
    P = (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?
    1
    T
    a
    m
    2
    Given: a = 5.0 m/s2; ` m = 100 kg
    W
    Find: 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?
    3
    T
    +
    a
    +
    W
    -
    4
    Upward (+)
    5
    Fnet = 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
    +
    5
    Fnet = T – mg
    W
    T – mg = ma
    -
    6
    T = mg + ma
    T = m(g + a)
    7
    T = (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?
    y
    FN
    x
    a
    m
    Wx
    30°
    W
  • 30. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?
    y
    FN
    x
    Fnet = Wx = W sin 30° Fnet= mg sin 30 °
    a
    mg sin 30 ° = ma
    Wx
    30°
    g sin 30 °= a
    9.8 m/s2sin 30 °= a
    a = 4.9 m/s2
    W
  • 31. Assignment
    Devise your own problem-solving strategy in solving problems in multiple-body systems in the application of the second law of motion?

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