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Let’s recall: How is the second law of motion represented mathematically? What does the unit “newton” mean?
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Application of the Second Law Single-Body Problems
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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
Indicate all given information and state what is to be found.
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Substitute all given quantities and solve for the unknown.
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From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.
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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.
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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?
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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 = ?
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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
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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
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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 = ?
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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
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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
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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
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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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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
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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
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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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