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

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.

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?