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# App of the 2nd law single body problems

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### App of the 2nd law single body problems

1. 1. Let’s recall:<br />How is the second law of motion represented mathematically?<br />What does the unit “newton” mean?<br />
2. 2. Application of the Second Law<br />Single-Body Problems<br />
3. 3. In which direction is the net force acting car A when it is moving east? <br />A<br /> In which direction is the net force acting car B when it is braking to a stop while moving east? <br />B<br />
4. 4. PROBLEM-SOLVING<br />TECHNIQUES<br />
5. 5. Mixed Up Recipe<br />
6. 6. <ul><li>Indicate all given information and state what is to be found.
7. 7. Substitute all given quantities and solve for the unknown.
8. 8. From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.
9. 9. Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = ma</li></ul>Read the problem carefully and then draw and label a rough sketch.<br /><ul><li>Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.
10. 10. Indicate a consistent positive direction along the continuous line of motion.</li></li></ul><li>Read the problem carefully and then draw and label a rough sketch.<br />2. Indicate all given information and state what is to be found.<br />3. Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.<br />4. Indicate a consistent positive direction along the continuous line of motion.<br />5. From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.<br />6. Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = ma<br />7. Substitute all given quantities and solve for the unknown.<br />
11. 11. Illustrative Example no.1<br />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?<br />m = 1000 kg<br />1<br />a = ?<br />v = 30.0 m/s<br />2<br />F = 4.0kN = 4000N<br />
12. 12. Draw a Free-Body Diagram<br />3<br />y<br />F = 4000 N<br />x<br />a<br />
13. 13. Rightward Positive<br />4<br />y<br />5<br />Fnet = - 4000 N<br />Fnet = ma<br />6<br />F = - 4000 N<br />Fnet<br />m<br />x<br />a = <br />-4000 N<br />1000 kg<br />a = ?<br />7<br />a = <br />a = -4.0 m/s2<br />
14. 14. Illustrative Example no.2<br />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? <br />m <br />F1<br />1<br />F2<br /> a <br />2<br /> a = ? <br />Given: <br />Find: <br />F1 =15.0 N<br />F2 =18.0 N<br />
15. 15. Draw a Free-Body Diagram<br />3<br />y<br />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?<br />Fnet = F1 + F2<br />x<br />a<br />m = 15.0 kg <br />
16. 16. Rightward positive<br />4<br />5<br />Fnet = F1 + F2<br />y<br />Fnet = 15.0N + 18.0N = 33.0N<br />Fnet = ma<br />6<br />Fnet<br />= +<br />Fnet<br />m<br />a = <br />x<br />a <br />= +<br />m = 15.0 kg <br />33.0 N<br />15.0 kg<br />7<br />a = <br />a = ?<br />a = 2.20 m/s2, Right <br />a = +2.20 m/s2<br />
17. 17. Illustrative Example no.3<br />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?<br />1<br />T<br />a<br />m<br />Given: a = 2.5 m/s2; T = 9600 N<br />2<br />Find: m = ?<br />W<br />
18. 18. Draw a Free-Body Diagram<br />3<br />y<br />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?<br />T <br />a<br />x<br />W <br />Weight = mass x acceleration due to gravity<br />W = mg <br />
19. 19. Upward, positive<br />4<br />5<br />Fnet = T – mg<br />T – mg = ma<br />y<br />6<br />T = ma + mg<br />= m(a + g)<br />T<br />(a + g)<br />T <br /> = +<br />m = <br />a <br /> = +<br />9600 N<br />2.5 m/s2 + 9.8 m/s2<br /> m = <br />7<br />x<br />m = ?<br />9600 N<br />12.3 m/s2<br /> m = <br />W = - mg <br />m = 780 kg <br />
20. 20. Solve the following<br />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?<br />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?<br />
21. 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?<br />a<br />1<br />P<br />m<br />f<br />2<br />Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 N<br />Find: P = ?<br />
22. 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?<br />Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 N<br />Find: P = ?<br />+<br />a<br />3<br />f<br />-<br />Free-body diagram<br />P<br />+<br />4<br />Rightward +<br />5<br />Fnet = P – f <br />
23. 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?<br />Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 N<br />Find: P = ?<br />5<br />Fnet = P – f <br />P = 24.0 N + 20.0 N<br />P – f = ma <br />6<br />Equate Fnet to ma, Fnet = ma<br />P = 44.0 N <br />P = ma + f <br />Derive equation to find the unknown<br />7<br />P = (6.0 kg)( 4.0 m/s2)+ 20.0 N <br />
24. 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?<br />1<br />T<br />a<br />m<br />2<br />Given: a = 5.0 m/s2; ` m = 100 kg<br />W<br />Find: T = ?<br />
25. 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?<br />3<br />T<br />+<br />a<br />+<br />W<br />-<br />4<br />Upward (+)<br />5<br />Fnet = T – W = T – mg <br />
26. 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?<br />T<br />+<br />a<br />+<br />5<br />Fnet = T – mg <br />W<br />T – mg = ma<br />-<br />6<br />T = mg + ma<br />T = m(g + a)<br />7<br />T = (100 kg)(9.8 m/s2+ 5.0 m/s2) <br />T = 100kg(14.8 m/s2)<br />= 1480 kg<br />
27. 27. Solve in paper number 4.<br />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 ?<br /> Ans. (a)F = 12.0 N; (b) 30,0 N and F =90.0 N <br />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. <br /> Ans. (a) 0; (b) -3.8 m/s2; (c) + 3.2 m/s2<br />
28. 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?<br />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?<br />
29. 29. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?<br />y<br />FN<br />x<br />a<br />m<br />Wx<br />30°<br />W<br />
30. 30. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?<br />y<br />FN<br />x<br />Fnet = Wx = W sin 30° Fnet= mg sin 30 °<br />a<br />mg sin 30 ° = ma<br />Wx<br />30°<br />g sin 30 °= a<br />9.8 m/s2sin 30 °= a<br />a = 4.9 m/s2<br />W<br />
31. 31. Assignment<br />Devise your own problem-solving strategy in solving problems in multiple-body systems in the application of the second law of motion?<br />